Question

The mean weight of U.S. men of age 30–39 years old is 191.802 pounds. Using the provided sample data file, conduct a one-sample ?-test of a mean to test whether the mean weight of men of age 30–39 years old who smoke daily is lower than 191.802.

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

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:

Data:

170	200	165	180	190	210	155	190	180	180	180	160	160	170	176	185	170	180	180
180	180	200	190	165	200	160	240	280	190	170	165	180	240	160	170	180	175	190
170	280	180	185	220	181	240	160	230	200	203	230	170	180	180	260	170	220	175
150	195	180	187	145	140	145	165	190	190	190	186	180	180	180	215	195	230	145
180	210	195	200	175	147	170	175	150	200	165	165	228	165	220	270	185	160	160
173	185	270	240	130	215	160	250	160	175	155	200	130	170	165	160	185	180	140
140	180	175	175	250	145	220	210	220	190	165	200	230	160	170	165	180	180	175
180	230	190	240	185	210	155	185	160	165	155	160	180	235	180	190	170	165	135
160	240	190	165	195	260	250	180	220	145	145	190	150	160	160	138	190	240	180
210	190	170	180	185	270	200	190	150	147	190	165	200	180	185	230	240	190

Answer 1

We can solve it via simple R-commands. Putting the data in a single row in an excel file, we can import the data in R as below.

> library(readxl)
> dat <- read_excel("data.xls", col_names = FALSE)

We have the sample mean as below.

> mean(dat$X__1)
[1] 186.3016

Hence, the sample mean is 186.3.

The standard deviation of the given distribution can be found as below.

> sd(dat$X__1)
[1] 30.75692

The standard error of mean is $\dpi{80} se = \frac{s}{\sqrt{n}}$ , where s is the sample standard deviation, and n is the number of elements. Hence, we have $\dpi{80} se = \frac{30.75692}{\sqrt{189}}$ or $\dpi{80} se = 2.237236731$ or $\dpi{80} se = 2.24$ (approx).

The standard error is hence 2.24.

For the hypothesis that $\dpi{80} H_0 : \bar{x} = 191.802$ and $\dpi{80} H_1 : \bar{x} < 191.802$ , the z-statistic is $\dpi{80} z = \frac{\bar{x} - 191.802}{se}$ or $\dpi{80} z = \frac{186.3 - 191.802}{2.24}$ or $\dpi{80} z = -2.45625$ .

The test will be one tailed. For a 5% alpha level, we have $\dpi{80} z_\alpha = z_{0.05} = -1.644854$ . As $\dpi{80} z < z_\alpha$ , we may reject the null $\dpi{80} H_0 : \bar{x} = 191.802$ , and accept $\dpi{80} H_1 : \bar{x} < 191.802$ .

Hence, the sample mean is indeed less than 191.802.