Question

Suppose a survey of 100 smokers, conducted by the Department of Health, suggestd that the mean number of ciggrettes a person smokes in a day in smokelandia (Y) is 2.72 and the standard deviation is 0.58. The Department of Health is concerned about the results ofthe survey and wants to test whether the eman number of cigarettes a person smokes in a day is 2.51 or not.

The test statistic associated with the above test is __________. (Round your answer to two decimal places.)
If the Department of Health uses a 5% significance level, the test statistic suggests that we ______ the hypothesis that the mean number of cigarettes a person smokes in a day is 2.51.

Answer 1

The test statistic associated with the above test is 3.62 . (Round your answer to two decimal places.)

If the Department of Health uses a 5% significance level, the test statistic suggests that we reject the hypothesis that the mean number of cigarettes a person smokes in a day is 2.51.

Step-by-step explanation:

The question involves conducting a hypothesis test to determine if the mean number of cigarettes a person smokes in a day in Smokelandia is different from 2.51. The sample mean (Y) is given as 2.72, with a standard deviation (σ) of 0.58, and the sample size (n) is 100 smokers. To find the test statistic, we would use the following formula for a one-sample z-test:

Z = (Y - μ) / (σ / √n)

Where:

Y = sample mean = 2.72

μ = hypothesized population mean = 2.51

σ = population standard deviation = 0.58

n = sample size = 100

Plugging the values into the formula, we get:

Z = (2.72 - 2.51) / (0.58 / √100)

Z = 0.21 / (0.58 / 10)

Z = 0.21 / 0.058

Z ≈ 3.62 (rounded to two decimal places)

Given a 5% significance level, the critical z-values for a two-tailed test are approximately -1.96 and +1.96. Since the calculated test statistic is greater than the positive critical value, we would reject the null hypothesis, which suggests that the mean number of cigarettes a person smokes per day is not 2.51.