Question

The claim is a mean is 76 and you want to prove it is less. Test the hypothesis within a 5% level of significance. Your sample of 99 had a mean of 74.48 and standard deviation of 12.16. Level of difficulty = 1 of 3 Please format to 2 decimal places.

Answer 1

To test the hypothesis that the mean is less than 76, perform a one-sample t-test with a significance level of 5%. The test statistic is calculated as -1.952, which is less than the critical value of -1.660. Therefore, we reject the null hypothesis and conclude that there is evidence to suggest that the mean is less than 76.

Step-by-step explanation:

To test the hypothesis, we need to perform a one-sample t-test. The null hypothesis (H0) is that the mean is equal to 76, while the alternative hypothesis (Ha) is that the mean is less than 76. Here are the steps to perform the test:

1. Calculate the test statistic: t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))
2. Find the critical value for a one-tailed test at a significance level of 5%. Since the alternative hypothesis is that the mean is less than 76, the critical value will be the t-value corresponding to a 5% upper tail.
3. Compare the test statistic to the critical value. If the test statistic is less than the critical value, reject the null hypothesis. Otherwise, fail to reject the null hypothesis.

In this case, the test statistic can be calculated as t = (74.48 - 76) / (12.16 / sqrt(99)) = -1.952. The critical value for a one-tailed test at a significance level of 5% (df = 99) is -1.660. Since the test statistic is less than the critical value, we reject the null hypothesis and conclude that there is evidence to suggest that the mean is less than 76.