Question

A data set about speed dating includes "like" ratings of male dates made by the female dates. The summary statistics are n=196, x̅=6.87, s=1.99. Use a 0.05 significance level to test the claim that the population mean of such ratings is less than 7.00. 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.

What are the null and altemative hypotheses?

Answer 1

A)

Hypothesis:

$H_{1}:\mu <7.00$

Test Statistics:

$n=196$

$\bar{x} = 6.87$

$s=1.99$

   $z=\frac{\bar{x}-\mu }{\frac{\sigma }{\sqrt{n}}}$

   $=\frac{6.87-7}{0.1421}$

   $=-0.9148$

P -value is 0.180

It is calculated using normal dist table.

Here 0.05 < 0.180.

We accept the null hypothesis.

The population mean of such ratings is equal to 7.00

B) Hypothesis:

$H_{0}:\mu =0$

$H_{1}:\mu >0$

Test Statistics:

$n=36$

$\bar{x} = 0.1$

$\sigma =23.6$

   $z=\frac{\bar{x}-\mu }{\frac{\sigma }{\sqrt{n}}} = \frac{0.1-0}{3.933} = 0.025$

P-value is 0.490

It is calculated using normal dist table.

Here 0.01 < 0.490

Hence we accept the null hypothesis.

This concludes that with garlic treatment the mean change in LDL cholestrol is equal to zero.