Question

In: Statistics and Probability

2. A researcher believes that smoking worsens a person’s sense of smell. To test this, he...

2. A researcher believes that smoking worsens a person’s sense of smell. To test this, he takes a sample of 25 smokers and gives them a test of olfactory sensitivity. In this test, higher scores indicate greater sensitivity. For his sample, the mean score on the test is 15.1 with a standard deviation of 1.2. The researcher knows the mean score in the population is 15.5, but the population standard deviation is unknown.

(a) What are the null and alternative hypotheses in this study (stated mathematically)?

(b) Should the researcher use a one-tailed or a two-tailed test?

(c) Compute the appropriate test statistic for testing the hypothesis.

(d) Using α = 0.01, do you conclude that smoking affects a person’s sense of smell? Be sure to include a discussion of the critical value in your answer.

(e) What type of error might the researcher be making in part (d)?

Solutions

Expert Solution

(a) Ho:

Ha:   

Null hypothesis states that the mean score of olfactory sensitivity for smokers is same as population mean = 15.5

(b) This is one tailed test. The higher sensitivity score indicate greater sensitivity. Where as the researches believes that smoking worsens a persons sense of smell, hence as per his claim the sensitivity score should be less than population mean.

(c) Assuming that the data is normally distributed. Since the sample size is less than 30 and population standard deviation is not given we will conduct t test

(d) level of significance = 0.01

The t-critical value for a left-tailed test, for a significance level of α=0.01

tc = −2.492

Since the t statistics (-1.667) does not fall in the rejection area we fail to reject the Null hypothesis.

As we fail to reject the Null hypothesis, we do not have sufficient evidence to believe that smoking worsens a persons sense of smell.

(e) The researcher might be making Type II error. i.e. actually the hypothesis is false but as per data test concludes that we fail to reject the Ho.


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