Chapter 10: Problem 64
A random sample of \(n=12\) observations from a normal population produced \(\bar{x}=47.1\) and \(s^{2}=4.7 .\) Test the hypothesis \(H_{0}: \mu=48\) against \(H_{\mathrm{a}}: \mu \neq 48 .\) Use the Small-Sample Test of a Population Mean applet and a \(5 \%\) significance level.
Short Answer
Expert verified
Based on the small-sample t-test with a 5% significance level, we fail to reject the null hypothesis, meaning there is not enough evidence to support the claim that the population mean is different from 48.
Step by step solution
01
State the null and alternative hypotheses
Null hypothesis (\(H_0\)): \(\mu = 48\).
Alternative hypothesis (\(H_a\)): \(\mu \neq 48\).
02
Calculate the t-statistic
To calculate the t-statistic, use the formula:
\(t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}\)
where \(\bar{x}\) is the sample mean, \(\mu_0\) is the hypothesized population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size. In this case, we have:
\(\bar{x} = 47.1\)
\(\mu_0 = 48\)
\(s^2 = 4.7\) (so \(s = \sqrt{4.7} \approx 2.1679\))
\(n = 12\)
Plug these values into the formula:
\(t = \frac{47.1 - 48}{\frac{2.1679}{\sqrt{12}}} \approx -1.2551\)
03
Determine the critical t-value
Since we are given a 5% significance level, and we are performing a two-tailed test, we will use a significance level of \(\frac{5\%}{2} = 2.5\%\) for each tail.
Using a t-distribution table or an online calculator with degrees of freedom (\(df = n-1 = 11\)) and a significance level of \(2.5\%\), we find the critical t-values to be \(\pm 2.20098\).
04
Compare t-statistic to critical t-value and decide
Now we compare our t-statistic and the critical t-value:
\(-2.20098 < -1.2551 < 2.20098\)
Our t-statistic (\(-1.2551\)) is not in the critical region (outside the range \(-2.20098\) to \(2.20098\)). Therefore, we fail to reject the null hypothesis.
05
Conclusion
We fail to reject the null hypothesis at the 5% significance level. There is not enough evidence to support the alternative hypothesis that the population mean is different from \(48\).
