Question

In: Statistics and Probability

According to the Centers for Disease Control and Prevention (CDC), the mean weight of U.S. men...

According to the Centers for Disease Control and Prevention (CDC), the mean weight of U.S. men of age 30–39 years old is191.802 pounds. Using the provided sample data file, conduct a one-sample ?-z-test of a mean to test whether the mean weight of men of age 30–39 years old who smoke daily is lower than191.802.

Conduct the z-test at a significance level of α=0.05 to test the null hypothesis, H0:μ=191.802, against the alternative hypothesis, H1:μ<191.802, where ?μ represents the mean weight of U.S. men of age 30–39 who smoke daily. Assume that the population standard deviation of weight, σ, is 42.221 pounds.

The sample data provided contain weight results collected from a 2015 survey conducted by the National Center for Health Statistics for men of age 30–39 who smoke daily.

Click to download the data in your preferred format.

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Calculate the sample mean, ?⎯⎯⎯x¯, for this data. Give your answer precise to at least one decimal place.

x¯ =

pounds

Compute the standard deviation, SD, of the sampling distribution of the mean. (Some people call this the standard error.) Give your answer with precision to at least two decimal places.

SD =

Next, determine the z-statistic for this hypothesis test. Provide your answer with precision to two decimal places. Avoid rounding until the final step.

z =

Using software, compute the P-value for the z-statistic. Give your answer precise to at least three decimal places. You may find software manuals helpful.

P-value =

If the test requires that the results be statistically significant at a level of α=0.05, fill in the blanks, and complete the sentences that explain the test decision and conclusion.

The decision is to

reject

the null hypothesis. There

is

enough evidence at a significance level of α=0.05 that the mean weight for U.S. men of age 30–39 who smoke daily is less than 191.802 pounds.

Solutions

Expert Solution

Assumed data,
sample mean, x =191.3
number (n)=50 because not given in the data
Given that,
population mean(u)=191.802
standard deviation, σ =42.221
null, Ho: μ=191.802
alternate, H1: μ<191.802
level of significance, α = 0.05
from standard normal table,left tailed z α/2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 191.3-191.802/(42.221/sqrt(50)
zo = -0.084
| zo | = 0.084
critical value
the value of |z α| at los 5% is 1.645
we got |zo| =0.084 & | z α | = 1.645
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value : left tail - ha : ( p < -0.084 ) = 0.466
hence value of p0.05 < 0.466, here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=191.802
alternate, H1: μ<191.802
test statistic: -0.084
critical value: -1.645
decision: do not reject Ho
p-value: 0.466
we do not have enough evidence to support the claim that the mean weight of men of age 30–39 years old who smoke daily is lower than 191.802.


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