Use the data given to test the following hupotheses. Assume the data are normally distributed in the population.
Ho: M=7.48 Ha:m>7.48
xbar = 6.91, n=24, sigma=1.21,alpha=.01
Since the sample size is small, you can use a t-test:
t = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)
With your data:
t = (6.91 - 7.48)/(1.21/√24)
Finish the calculation.
Check a t-table using n - 1 for degrees of freedom for a one-tailed test at .01 level of significance (the alternate hypothesis is showing a specific direction). If the test statistic exceeds the critical value from the table, reject the null. If the test statistic does not exceed the critical value from the table, do not reject the null.
You can draw your conclusions from there.
I hope this will help get you started.
-2.28
To test the given hypotheses, we can use a one-sample Z-test.
First, let's define the null hypothesis (Ho) and the alternative hypothesis (Ha):
Ho: M = 7.48 (Mean is equal to 7.48)
Ha: M > 7.48 (Mean is greater than 7.48)
Next, we calculate the standard error (SE) which measures the average difference between the sample mean and the population mean. The formula for SE is:
SE = sigma / sqrt(n)
Given data:
xbar = 6.91 (sample mean)
n = 24 (sample size)
sigma = 1.21 (standard deviation)
SE = 1.21 / sqrt(24) = 0.247
Now, we calculate the test statistic (Z). The formula for Z-test is:
Z = (xbar - M) / SE
Z = (6.91 - 7.48) / 0.247 = -2.306
To determine the critical value of Z for the given significance level (alpha = 0.01), we consult the Z-table or use statistical software. For a one-tailed test with alpha = 0.01, the critical value is approximately 2.33.
Since -2.306 is less than -2.33, it falls in the rejection region. Thus, we can reject the null hypothesis in favor of the alternative hypothesis.
In conclusion, based on the given data, there is evidence to suggest that the mean (M) is greater than 7.48.