In a study on speed dating females rated the “attractiveness” of males. In a ran

Question

In a study on speed dating females rated the “attractiveness” of males. In a random sample of
187 female speed dating participants, on average men received a 6.14 on a scale from 1 to 10. It
is believed from the results of previous studies that these ratings are normally distributed with a
standard deviation of 1.73. Use a 0.01 significance level to test the claim that the population of
such ratings is less than 7.00.
a. Identify the null and alternative hypothesis
b. Calculate by hand the test statistic for this hypothesis test
c. Use Table A-2 to determine the necessary P-Value.
d. State the final conclusion which addresses the original claim.
e. Identify a major issue with using inferential methods with this mean attractiveness
rating. (Hint: Consider an individual’s attractiveness rating. Would someone get the
same rating from every other person?)

Explanation / Answer

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: > 7.0

Alternative hypothesis: < 7.0

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample z-test.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 0.1265

z = (x - ) / SE

z = 6.8

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a t statistic test statistic of 6.8. We use the z Distribution Calculator to find P(z > 6.8).

Thus the P-value in this analysis is less than 0.0001.

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we have to reject the null hypothesis.

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