Question

you wish to test the following claim ($h_a$) at a significance level of $\alpha = 0.10$.
$h_0: \mu = 71.1$
$h_a: \mu\
eq71.1$
you believe the population is normally distributed, but you do not know the standard deviation. you obtain the following sample of data:
| data |
| ---- |
| 71.5 |
| 112.9 |
| 96.6 |
| 70.7 |
what is the test statistic for this sample? (report answer accurate to 3 decimal places.)
test statistic =
what is the p - value for this sample? (report answer accurate to 3 decimal places.)
p - value =
the p - value is...
less than (or equal to) $\alpha$
greater than $\alpha$
this test statistic leads to a decision to...
reject the null
fail to reject the null
as such, the final conclusion is that...
the sample data support the claim that the population mean is not equal to 71.1.
there is not sufficient sample evidence to support the claim that the population mean is not equal to 71.1.

Explanation:

Step1: Calculate sample mean

The sample data is \(71.5, 112.9, 96.6, 70.7\). The sample - mean \(\bar{x}=\frac{71.5 + 112.9+96.6 + 70.7}{4}=\frac{351.7}{4}=87.925\).

Step2: Calculate sample standard deviation

The formula for the sample standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\).
\((71.5−87.925)^{2}=(-16.425)^{2}=269.880625\)
\((112.9−87.925)^{2}=(24.975)^{2}=623.750625\)
\((96.6−87.925)^{2}=(8.675)^{2}=75.255625\)
\((70.7−87.925)^{2}=(-17.225)^{2}=296.795625\)
\(\sum_{i = 1}^{4}(x_{i}-\bar{x})^{2}=269.880625 + 623.750625+75.255625 + 296.795625 = 1265.6825\)
\(s=\sqrt{\frac{1265.6825}{3}}\approx20.532\)

Step3: Calculate the t - test statistic

The formula for the t - test statistic in a one - sample t - test is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\), where \(\mu = 71.1\), \(\bar{x}=87.925\), \(s\approx20.532\), and \(n = 4\).
\(t=\frac{87.925 - 71.1}{20.532/\sqrt{4}}=\frac{16.825}{10.266}\approx1.64\)

Step4: Calculate the P - value

The degrees of freedom \(df=n - 1=4 - 1 = 3\). Since this is a two - tailed test, the P - value is \(2P(T>1.64)\) where \(T\) follows a t - distribution with \(df = 3\). Using a t - distribution table or a calculator, \(P(T>1.64)\) with \(df = 3\) is approximately \(0.119\), so the P - value \(=2\times0.119 = 0.238\)

Step5: Make a decision

Since \(\alpha=0.10\) and the P - value \(=0.238>0.10\), we fail to reject the null hypothesis.

Answer:

test statistic = \(1.640\)
P - value = \(0.238\)
The P - value is greater than \(\alpha\)
This test statistic leads to a decision to fail to reject the null
As such, the final conclusion is that there is not sufficient sample evidence to support the claim that the population mean is not equal to \(71.1\)