Question

A random sample of 18 women is taken and their heights were recorded. The heights (in inches) are:

60, 62, 63, 63, 63, 66, 66, 66, 66, 67, 67, 68, 68, 68, 69, 70, 71, 71

Assume that women’s height are normally distributed. Let μ be the mean height of all women and let p be the proportion of all women that are taller than 54 inches.

a) Test the hypothesis H0: μ = 54 against H1: μ > 54. Find the p-value and report your conclusion.

b) Test the hypothesis H0: p = 0.5 against H1: p > 0.5. Find the p-value and report your conclusion.

Answer 1

Assuming level of significance to be 5%

a)

One-Sample t-test

The sample mean is Xˉ=66.3333, the sample standard deviation is s=3.1249, and the sample size is n=18. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: μ =54 Ha: μ >54 This corresponds to a Right-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2a) Critical Value Based on the information provided, the significance level is α=0.05, and the degree of freedom is n-1=18-1=17. Therefore the critical value for this Right-tailed test is tc​=1.7396. This can be found by either using excel or the t distribution table. (2b) Rejection Region The rejection region for this Right-tailed test is t>1.7396 (3)Test Statistics The t-statistic is computed as follows: (4) The p-value The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true. In this case, the p-value is 0 (5) The Decision about the null hypothesis (a) Using traditional method Since it is observed that t=16.7451 > tc​=1.7396, it is then concluded that the null hypothesis is rejected. (b) Using p-value method Using the P-value approach: The p-value is p=0, and since p=0≤0.05, it is concluded that the null hypothesis is rejected. (6) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is greater than 54, at the 0.05 significance level.

b)

All women are taller than 54 inches.

One-Proportion Z test

The following information is provided: The sample size is N = 18, the number of favorable cases is X = 18 and the sample proportion is pˉ​=X/N​=18/18​=1, and the significance level is α=0.05 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: p =0.5 Ha: p >0.5 This corresponds to a Right-tailed test, for which a z-test for one population proportion needs to be used. (2a) Critical Value Based on the information provided, the significance level is α=0.05, therefore the critical value for this Right-tailed test is Zc​=1.6449. This can be found by either using excel or the Z distribution table. (2b) Rejection Region The rejection region for this Right-tailed test is Z>1.6449 (3) Test Statistics The z-statistic is computed as follows: (4) The p-value The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true. In this case, the p-value is p =P(Z>4.2426)=0 (5) The Decision about the null hypothesis (a) Using traditional method Since it is observed that Z=4.2426 > Zc​=1.6449, it is then concluded that the null hypothesis is rejected. (b) Using p-value method Using the P-value approach: The p-value is p=0, and since p=0≤0.05, it is concluded that the null hypothesis is rejected. (6) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population proportion p is greater than 0.5, at the 0.05 significance level.

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