1. A ___________ is a range of values that probably contains the population valu

Question

1. A ___________ is a range of values that probably contains the population value.

2. Setting the ______ to 0.05 means that over the long run the researcher is willing to be wrong 5% of the time.

Construct Confidence Intervals for the following (Note: show your calculations below)

3. Sample Mean = 5.2

Sample Standard Deviation = 0.7

Sample Size (n) = 157

Alpha Level = 0.05

4. Sample Mean = 1020

Population Standard Deviation = 50

Sample Size (n) = 329

Alpha Level = 0.05

5. The average income for a random sample (n = 500) of a particular community is $35,000, with a sample standard deviation of $200. What is the 95% interval estimate of the population?

6. In a sample of 200 freshmen at a state university, 40% report that they work at least 20 hours a week while in school. Estimate the proportion of all freshmen at the university working at least 20 hours per week. Develop your estimate on the basis of a 95% confidence interval.

Explanation / Answer

1) Confidence Interval

2) alpha

3)

Sample Mean = 5.2

Sample SD =0.7

Sample Size(n) = 157

Alpha = 0.05

compute the margin of error, based on one of the following equations.

Margin of error = Critical value * Standard deviation of statistic
Margin of error = Critical value * Standard error of statistic

Like in this case we will use t-statistic

To find the critical value, follow these steps.

df =degrees of freedom = n - 1 = 157 - 1 = 156

And cumulative/critical probability = 0.975

USea t-table or t value calculator we get t-score(crtiical value)= 1.975

Find standard error. The standard error (SE) of the mean is:

SE = s / sqrt( n ) = 0.7/sqrt(157)=0.05587

Compute margin of error (ME): ME = critical value * standard error = 0.05587 X 1.975 =0.1103

The range of the confidence interval is defined by the sample statistic + margin of error. And the uncertainty is denoted by the confidence level. Therefore, this 95% confidence interval is 5.2+0.1103

(4)

The standard error (SE) of the mean is:

SE = s / sqrt( n ) = 50/sqrt(329)= 2.7565

Z value is 1.96 for 95% confidence interval

Therefore

Compute margin of error (ME): ME = critical value * standard error = 1.96 X 2.7565 = 5.40

therefore

CI = 1020 + 5.40

I have done 4sub parts , for rest put the 5th question again

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