College

You wish to test the following claim \([tex]H_a[/tex]\) at a significance level of \([tex]\alpha=0.01[/tex]\).

\[
\begin{align*}
H_0: & \quad \mu_1 = \mu_2 \\
H_a: & \quad \mu_1 < \mu_2
\end{align*}
\]

You obtain the following two samples of data.

**Sample #1**

\[
\begin{array}{|r|r|r|r|}
\hline
72 & 63.2 & 58.4 & 73.2 \\
\hline
62.7 & 69.2 & 76.6 & 73.5 \\
\hline
73.5 & 70.6 & 83 & 57.5 \\
\hline
86.9 & 78.4 & 77.5 & 70 \\
\hline
78.4 & 59.3 & 76.6 & 78.4 \\
\hline
82.6 & 74.5 & 94.2 & 69.6 \\
\hline
75.4 & 86.9 & 71.3 & 75.1 \\
\hline
77.5 & 59.3 & 96.2 & 71 \\
\hline
94.2 & 83.7 & 78.4 & 77.8 \\
\hline
67.4 & 71.6 & 99.1 & 97.5 \\
\hline
96.2 & 86.1 & 67 & 90.5 \\
\hline
65.2 & 51.4 & 66.5 & 91.1 \\
\hline
82.6 & 68.1 & 104.4 & 69.6 \\
\hline
66.5 & 97.5 & 85.2 & 91.8 \\
\hline
\end{array}
\]

**Sample #2**

\[
\begin{array}{|r|r|r|r|}
\hline
62.9 & 64 & 96.5 & 105.8 \\
\hline
66.7 & 64.6 & 68.2 & 75.6 \\
\hline
97.3 & 59.5 & 71.4 & 69.6 \\
\hline
76 & 82.1 & 91.7 & 64 \\
\hline
74.3 & 78.8 & 68.7 & 71.8 \\
\hline
67.2 & 105.8 & 88 & 67.2 \\
\hline
75.6 & 64 & 85.6 & 77.2 \\
\hline
57 & 69.6 & 56 & 84.2 \\
\hline
79.6 & 85.6 & 72.7 & 64 \\
\hline
102.6 & 90.6 & 79.2 & 89 \\
\hline
100.2 & 88 & 93 & 76 \\
\hline
75.6 & 64 & 67.2 & 95.7 \\
\hline
59.5 & 56 & 95 & 88 \\
\hline
64 & 69.2 & 95.7 & 99.2 \\
\hline
74.3 & & & \\
\hline
\end{array}
\]

What is the test statistic for this sample? (Report answer accurate to three decimal places.)

Test statistic = \([tex]\square[/tex]\)

What is the p-value for this sample? For this calculation, use the degrees of freedom reported from the technology you are using. (Report answer accurate to four decimal places.)

p-value = \([tex]\square[/tex]\)

Answer :

We begin by noting that we wish to test

[tex]$$
\begin{aligned}
H_0&:\mu_1 = \mu_2, \\
H_a&:\mu_1 < \mu_2,
\end{aligned}
$$[/tex]

with a significance level of [tex]$\alpha=0.01$[/tex]. Two independent samples of data were collected. Their sizes, means, and variances (calculated with [tex]$n-1$[/tex] degrees of freedom for the sample variance) are as follows:

- For Sample \#1 with [tex]$n_1=56$[/tex], the sample mean is [tex]$\bar{x}_1 \approx 77.1768$[/tex] and the sample variance is [tex]$s_1^2 \approx 144.9905$[/tex].

- For Sample \#2 with [tex]$n_2=57$[/tex], the sample mean is [tex]$\bar{x}_2 \approx 77.7333$[/tex] and the sample variance is [tex]$s_2^2 \approx 189.4001$[/tex].

Because the sample variances differ, we use the two-sample [tex]$t$[/tex]-test for unequal variances (Welch’s [tex]$t$[/tex]-test). The test statistic is calculated using

[tex]$$
t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}.
$$[/tex]

Substituting the numbers into the formula gives

[tex]$$
t \approx \frac{77.1768 - 77.7333}{\sqrt{\frac{144.9905}{56} + \frac{189.4001}{57}}} \approx -0.2289.
$$[/tex]

Rounded to three decimal places, this yields

[tex]$$
t \approx -0.229.
$$[/tex]

Next, we determine the degrees of freedom (using the Welch–Satterthwaite approximation)

[tex]$$
df = \frac{\left(\dfrac{s_1^2}{n_1} + \dfrac{s_2^2}{n_2}\right)^2}{\frac{\left(\dfrac{s_1^2}{n_1}\right)^2}{n_1-1} + \frac{\left(\dfrac{s_2^2}{n_2}\right)^2}{n_2-1}}.
$$[/tex]

The computed value is approximately [tex]$df \approx 109.5487$[/tex]. (For our one-tailed test, the exact degrees of freedom used in the technology are taken into account.)

Because the alternative hypothesis is [tex]$H_a: \mu_1 < \mu_2$[/tex], we need the one-tailed [tex]$p$[/tex]-value corresponding to the test statistic. For a negative [tex]$t$[/tex] statistic, the one-tailed [tex]$p$[/tex]-value is given by the cumulative distribution function value

[tex]$$
p\text{-value} = P(T \le t),
$$[/tex]

where [tex]$T$[/tex] follows a [tex]$t$[/tex]-distribution with approximately [tex]$109.5487$[/tex] degrees of freedom. For [tex]$t \approx -0.2289$[/tex], the one-tailed [tex]$p$[/tex]-value computed comes out to

[tex]$$
p\text{-value} \approx 0.4097.
$$[/tex]

Thus, our final answers are:

[tex]$$
\text{Test statistic} = -0.229 \quad \text{(to three decimal places)},
$$[/tex]

[tex]$$
\text{p-value} = 0.4097 \quad \text{(to four decimal places)}.
$$[/tex]

Since the [tex]$p$[/tex]-value is much larger than the significance level [tex]$\alpha=0.01$[/tex], we do not have sufficient evidence to reject the null hypothesis.