Question

A data set about speed dating includes​ "like" ratings of male dates made by the female dates. The summary statistics are nequals189​, x overbarequals7.58​, sequals1.93. Use a 0.05 significance level to test the claim that the population mean of such ratings is less than 8.00. Assume that a simple random sample has been selected. Identify the null and alternative​ hypotheses, test​ statistic, P-value, and state the final conclusion that addresses the original claim.

Answer 1

Answer with explanation:

Given : Sample size $n= 189$

Sample mean : $\overline{x}=7.58$

Sample standard deviation : $s=1.93$

Let $\mu$ be the population mean of "like" ratings of male dates made by the female dates.

As per question ,

Null hypothesis : $\mu\geq8.00$

Alternative hypothesis : $\mu$ , It means the test is a one-tailed t-test. ( we use t-test when population standard deviation is unknown.)

Test statistic:

$t_{stat}=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}\\\\=\dfrac{7.58-8.00}{\dfrac{1.93}{\sqrt{189}}}=-2.99$

For 0.05 significance and df =188 (df=n-1) p-value = .001582. [By t-table]

Since  .001582< 0.05

Decision: p-value < significance level , that means there is statistical significance, so we reject the null hypothesis.

Conclusion : We support the claim at 5% significance that t the population mean of such ratings is less than 8.00.