A pharmaceutical company makes tranquilizers. It is assumed that the distributio

Question

A pharmaceutical company makes tranquilizers. It is assumed that the distribution for the length of time they last is approximately normal. Researchers in a hospital used the drug on a random sample of 9 patients. The effective period of the tranquilizer for each patient (in hours) was as follows: 2.7; 2.9; 3.0; 2.3; 2.3; 2.2; 2.8; 2.1; and 2.4.

NOTE: If you are using a Student's t-distribution, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

Part (a)

(i) Round your answer to two decimal places.

x =

(rounded to two decimal places)

(ii) Round your answer to two decimal places.

sx =

(rounded to two decimal places)

(iii) Enter an exact number as an integer, fraction, or decimal.

n =

(iv) Enter an exact number as an integer, fraction, or decimal.

n ? 1 =

Part (b)

In words, define the random variable X.

the number of patients that were given a tranquilizerthe number of tranquilizers the hospital used for each patient    time, in minutes, of the effectiveness of a tranquilizertime, in hours, of the effectiveness of a tranquilizer

Part (c)

In words, define the random variable

X.

the average number of tranquilizers the pharmaceutical company dispenses to hospitalsthe average length, in hours, of the effectiveness period of tranquilizers    the average length, in minutes, of the effectiveness period of tranquilizersthe average number of tranquilizers used for each patient

Part (d)

Which distribution should you use for this problem? (Enter your answer in the form z or tdf where df is the degrees of freedom.)

Explain your choice.

The standard normal distribution should be used because the sample standard deviation is known.The Student's t-distribution should be used because the sample standard deviation is known and the sample size is small.    The Student's t-distribution should be used because the sample size is small.The standard normal distribution should be used because the population standard deviation is known.

Part (e)

Construct a 95% confidence interval for the population mean length of time.(i) State the confidence interval. (Round your answers to two decimal places.)

  ,

(ii) Sketch the graph.

(iii) Calculate the error bound. (Round your answer to two decimal places.)

Part (f)

What does it mean to be "95% confident" in this problem?

This means that we are 95% confident that the average length of effectiveness of tranquilizers in the sample of 9 people is between the interval values.We are 95% confident that the effectiveness of a tranquilizer lies between the interval values.    This means that the chances of a tranquilizer being effective is 95%.This means that if intervals are created from repeated samples, 95% of them will contain the true population average length of effectiveness of tranquilizers.

Suppose that insurance companies did a survey. They randomly surveyed 410 drivers and found that 340 claimed they always buckle up. We are interested in the population proportion of drivers who claim they always buckle up.

NOTE: If you are using a Student's t-distribution, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

Part (a)

(i) Enter an exact number as an integer, fraction, or decimal.

x =

(ii) Enter an exact number as an integer, fraction, or decimal.

n =

(iii) Round your answer to four decimal places.

p' =

(rounded to four decimal places)

Part (b)

In words, define the random variables X and P'.

X is the proportion of people in the sample who claim they buckle up, and P' is the number of people who buckle up.X is the number of people who claim they buckle up, and P' is the proportion of people in the sample who buckle up.    X is the proportion of people in the sample who do not buckle up, and P' is the number of people who do not buckle up.X is the number of people who do not buckle up, and P' is the proportion of people in the sample who do not buckle up.

Part (c)

Which distribution should you use for this problem? (Round your answer to four decimal places.)

P' ~

  ,

Explain your choice.The normal distribution should be used because we are interested in proportions and the sample size is large.The binomial distribution should be used because there are two outcomes, buckle up or do not buckle up.    The Student's t-distribution should be used because

? 10, which implies a small sample.The Student's t-distribution should be used because we do not know the standard deviation.

Part (d)

Construct a 95% confidence interval for the population proportion who claim they always buckle up.(i) State the confidence interval. (Round your answers to four decimal places.)

  ,

(ii) Sketch the graph.

(iii) Calculate the error bound. (Round your answer to four decimal places.)

Part (e)

If this survey were done by telephone, list three difficulties the companies might have in obtaining random results.

The individuals in the sample may not accurately reflect the population.Individuals may choose to participate or not participate in the phone survey.Individuals may not tell the truth about buckling up.Only people over the age of 24 would be included in the survey.Children who do not answer the phone will not be included.

An article regarding interracial dating and marriage recently appeared in a newspaper. Of the 1716 randomly selected adults, 307 identified themselves as Latinos, 322 identified themselves as blacks, 255 identified themselves as Asians, and 779 identified themselves as whites. Among Asians, 79% would welcome a white person into their families, 71% would welcome a Latino, and 66% would welcome a black person.

NOTE: If you are using a Student's t-distribution, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

Part (a)

Construct the 95% confidence intervals for the three Asian responses. (Round your answers to four decimal places.)

Part (b)

Even though the three point estimates are different, do any of the confidence intervals overlap? Which? (Select all that apply.)

Yes, the intervals for whites and blacks overlap.No confidence intervals overlap.Yes, the intervals for whites and Latinos overlap.Yes, the intervals for Latinos and blacks overlap.Yes, all three intervals overlap.

Part (c)

For any intervals that do overlap, in words, what does this imply about the significance of the differences in the true proportions?

Intervals that overlap do not give any indication about the difference of the true population proportion.The confidence intervals for white and Latino overlap, as do those for black and Latino, which means that there is no significant difference in their proportions.    The confidence intervals for black and Latino overlap, and the overlapping region contains the black and Latino sample proportions; therefore, there is no evidence of a significant difference between their true proportions. The confidence intervals for white and Latino overlap, but the overlapping region does not contain the white or Latino sample proportions; therefore, there is evidence of a significant difference between their true proportions.There are no confidence intervals that overlap, implying that there is a significant difference in their population proportions.

Part (d)

For any intervals that do not overlap, in words, what does this imply about the significance of the differences in the true proportions?

Intervals that do not overlap do not give any indication about the difference of the true population proportion.The intervals for black and white do not overlap, which means that there is no significant difference in their population proportions.    The intervals for black and white do not overlap, which means that there is a significant difference in their population proportions.There are no intervals that do not overlap, implying that there is not a significant difference in their population proportions.

A sample of 11 small bags of the same brand of candies was selected. Assume that the population distribution of bag weights is normal. The weight of each bag was then recorded. The mean weight was 3 ounces with a standard deviation of 0.13 ounces. The population standard deviation is known to be 0.1 ounce.

NOTE: If you are using a Student's t-distribution, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

Part (a)

Find the following. (Round your answers to two decimal places.)(i)    

0.1

(iii)    

sx =

Part (b)

In words, define the random variable X.

Part (c)

In words, define the random variable

X.

Part (d)

Which distribution should you use for this problem? (Round your answers to three decimal places.)

X ~

? U H Exp B N

  ,

Explain your choice.

The Student's t-distribution should be used because the sample size is small.The standard normal distribution should be used because the population standard deviation is known.    The standard normal distribution should be used because the sample standard deviation is known.The Student's t-distribution should be used because the sample standard deviation is given.

Part (e)

Construct a 90% confidence interval for the population mean weight of the candies.(i) State the confidence interval. (Round your answers to three decimal places.)

  ,

(ii) Sketch the graph.

(iii) Calculate the error bound. (Round your answer to three decimal places.)

Part (f)

Construct a 98% confidence interval for the population mean weight of the candies.(i) State the confidence interval. (Round your answers to three decimal places.)

  ,

(ii) Sketch the graph.

(iii) Calculate the error bound. (Round your answer to three decimal places.)

Part (g)

In complete sentences, explain why the confidence interval in part (f) is larger than the confidence interval in part (e).

The confidence interval in part (f) is larger than the confidence interval in part (e) because a small sample size is being used.The confidence interval in part (f) is larger than the confidence interval in part (e) because a larger level of confidence increases the error bound, making the interval larger.    The confidence interval in part (f) is larger than the confidence interval in part (e) because the mean weight changes for each sample.The confidence interval in part (f) is larger than the confidence interval in part (e) because the population standard deviation changes for each sample.

Part (h)

In complete sentences, give an interpretation of what the interval in part (f) means.

We are 98% confident that the true population mean weight of all small bags of candies is between these values.We are 98% confident that a small bag of candies weighs between these values.    We are 98% confident that the mean weight of the sample of 11 small bags of candies is between these values.There is a 98% chance that a small bag of candies weighs between these values.

Explanation / Answer

since n<30 use t distribution

code in R

effectiveperiod <- c(2.7,2.9, 3.0,2.3, 2.3, 2.2, 2.8, 2.1, 2.4)
mean(effectiveperiod)
sd(effectiveperiod)

Output:

mean(effectiveperiod)
[1] 2.522222
> sd(effectiveperiod)
[1] 0.3308239
>

(i) Round your answer to two decimal places.

(i) Round your answer to two decimal places.

x =2.52

(rounded to two decimal places)

(ii) Round your answer to two decimal places.

sx =0.33

(rounded to two decimal places)

(iii) Enter an exact number as an integer, fraction, or decimal.

n =9

iv)

n-1=9-1=8

The effective period of the tranquilizer for each patient (in hours)

Part (c)

In words, define the random variable

X.

average length, in hours, of the effectiveness period of tranquilizers  

Part (d)

Which distribution should you use for this problem? (Enter your answer in the form z or tdf where df is the degrees of freedom.)

t distribution

since n=9,n<30 and degrees of freedom=n-1=9-1=8

.The Student's t-distribution should be used because the sample standard deviation is known and the sample size is small.  

Part (e)

Construct a 95% confidence interval for the population mean length of time.(i) State the confidence interval. (Round your answers to two decimal places.)

t.test(effectiveperiod)

One Sample t-test

data: effectiveperiod
t = 22.872, df = 8, p-value = 1.416e-08
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
2.267928 2.776516
sample estimates:
mean of x
2.522222

95 percent confidence interval:
2.27 and 2.78

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