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

Null hypothesis:

Alternative hypothesis:

The p value for this case is significantly lower than the significance level of 0.1 so then we can conclude that we have enough evidence to conclude that the true mean for the ratings is lower then 7.00 at 10% of significance

We have the following info given from the problem

represent the sample mean

represent the sample deviation

represent the sample size

We want to conduct an hypothesis to check the following hypothesis:

System of hypothesis

Null hypothesis:

Alternative hypothesis:

Statistic

The statistic in order to check this hypothesis since we don't know the population deviation is given by:

And replacing we got:

Now we can find the degrees of freedom given by:

P value

Now we can find the p value taking in count that we have a left tailed test:

Final conclusion

The p value for this case is significantly lower than the significance level of 0.1 so then we can conclude that we have enough evidence to conclude that the true mean for the ratings is lower then 7.00 at 10% of significance

Given : Sample size

Sample mean :

Sample standard deviation :

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

As per question ,

Null hypothesis :

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

Test statistic:

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.

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