Null hypothesis H0: μ "\\leq" 7.5
Alternative hypotheses H1: μ > 7.5
The level of significance can be considered as 0.05.
The test statistic for one sample t test is:
"t = \\frac{x - \u03bc}{\\frac{\u03c3}{\\sqrt{n}}} \\\\\n\n= \\frac{8.5 - 7.5}{\\frac{1.5}{\\sqrt{30}}} \\\\\n\n= \\frac{1.0}{0.274} \\\\\n\n= 3.65"
The degrees of freedom can be calculated as:
n – 1 = 30 – 1
= 29
Determine the p-value of the test from standard normal table as below:
"P(Z>z) = 1-P(Z<z) \\\\\n\n= 1 -P(Z< 3.65) \\\\\n\n= 1 - 0.9998 \\\\\n\n= 0.0002 \u2248 0"
The decision is to reject the null hypothesis if p-value is less than the considered level of significance. Here, p-value is less than the significance level. The researcher will reject the null hypothesis. Hence, there is sufficient evidence to say that the population mean is less than 7.5.
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