Given a sample mean of 45.1 and sample standard deviation 4.6, test the claim that the population mean equals 46 at α = 0.05 with n = 158.

What we are checking

You are testing whether the population mean is exactly 46, using a large sample where the population standard deviation is unknown. That calls for a one-sample $t$ test (two-tailed).

Set up the hypotheses

Because the claim is “equal to 46,” the equality goes in the null hypothesis:

• $H_0: \mu = 46$
• $H_a: \mu \ne 46$

Compute the test statistic

First find the standard error:

$$SE=\frac{s}{\sqrt{n}}=\frac{4.6}{\sqrt{158}}\approx \frac{4.6}{12.57}\approx 0.366$$

Now compute the $t$ statistic:

$$t=\frac\bar{x}-\mu_0{SE}=\frac{45.1-46}{0.366}\approx \frac{-0.9}{0.366}\approx -2.46$$

Degrees of freedom: $df=n-1=157$.

p-value and decision at $\alpha=0.05$

This is two-tailed, so we look at the two-tailed p-value for $t=-2.46$ with $df=157$:

• $p \approx 0.015$

Decision rule: reject $H_0$ if $p<0.05$.

Since $0.015<0.05$, reject $H_0$.

What the result means (in context)

There is statistically significant evidence that the population mean is different from 46. Because the sample mean is 45.1, the data suggest the true mean is likely less than 46.

(Optional check via confidence interval) A 95% CI is

$$\bar{x} \pm t_{0.975,157}\cdot SE \approx 45.1 \pm 1.975(0.366) \approx 45.1 \pm 0.723$$

So the CI is approximately $(44.38,\ 45.82)$, which does not include 46.