8.1 A job placement director claims that mean starting salary for nurses is $25

Question

8.1

A job placement director claims that mean starting salary for nurses is $25 per hour. A random sample of 10 nurses salaries has a mean $21.6 and a standard deviation of $4.7 per hour. Is there enough evidence to reject the directors claim at alpha=0.01?

If using the above data with sample mean $21.6 and a standard deviation of $.47. What is your answer now? What can you conclude about the role of noise (standard deviation) in statistical testing?

8.2

Suppose that 95% CI for the mean of a large sample was computed and equaled [10.15,10.83] . What will be your decision about the hypothesis Ho: mu=10 vs Ha: mu does not =10 at 5% level of significance? At 10% level? At 1% level?

8.3

The sports research lab studies the effects of swimming on max volume of oxygen uptake. For 8 volunteers the maximal oxygen uptake was measured before and after the 6 week swimming program. The results:

Before 2.1          3.3          2.0          1.9          3.5          2.2          3.21        2.4

After     2.7          3.5          2.8          2.3          3.2          2.1          3.6          2.9

Is there evidence that the swimming program has increased the maximal oxygen uptake?

8.4

Visitors to the electronics website rated their satisfaction with 2 models of printers/scanners, on the scale of 1 to 5. The following were obtained:

                                N             mean    st. dev.

Model A               31           3.6          1.5

Model B               65           4.2          0.9

At a level of 5%, test the hypothesis that both printers would have the same average rating in the general population, that is Ho: mu a =mu b. Also, calculate the 95% confidence interval for the mean difference mu a – mu b.

8.5

You are studying yield of a new variety of tomato, In the past, yields of similar types of tomato have shown a standard deviation of 8.5 lb. per plant. You would like to design a study that will determine the average yield within a 90% error margin of plus or minus 2 lbs. How many plants should you sample?

Explanation / Answer

Solution:-

8.1)

a)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: = 25

Alternative hypothesis: 25

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 1.486

DF = n - 1 = 10 - 1

D.F = 9

t = (x - ) / SE

t = - 2.29

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the t statistic having 9 degrees of freedom is less than - 2.29 or greater than 2.29.

Thus, the P-value = 0.1724

Interpret results. Since the P-value (0.08) is greater than the significance level (0.01), we cannot reject the null hypothesis.

b)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: = 25

Alternative hypothesis: 25

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 0.1486

DF = n - 1 = 10 - 1

D.F = 9

t = (x - ) / SE

t = - 22.8

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the t statistic having 9 degrees of freedom is less than - 22.8 or greater than 22.8

Thus, the P-value = less than 0.0001

Interpret results. Since the P-value (almost 0) is less than the significance level (0.01), we have to reject the null hypothesis.

As the standard deviation of the data set decreases the significance of the test increases.

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