Question

you wish to test the following claim (hₐ) at a significance level of α = 0.10. h₀: μ = 61.5 hₐ: μ > 61.5 you believe the population is normally distributed, but you do not know the standard deviation. you obtain the following sample of data: data 96.4 95.7 73.3 92.8 76.8 99.7 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) α greater than α 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 greater than 61.5 there is not sufficient sample evidence to support the claim that the population mean is greater than 61.5

Explanation:

Step1: Calculate sample mean $\bar{x}$

$\bar{x}=\frac{96.4 + 95.7+73.3+92.8+76.8+99.7}{6}=\frac{534.7}{6}\approx89.117$

Step2: Calculate sample standard - deviation $s$

First, calculate the differences from the mean:
$d_1=96.4 - 89.117 = 7.283$, $d_2=95.7 - 89.117 = 6.583$, $d_3=73.3 - 89.117=-15.817$, $d_4=92.8 - 89.117 = 3.683$, $d_5=76.8 - 89.117=-12.317$, $d_6=99.7 - 89.117 = 10.583$
Then, square the differences:
$d_1^2 = 7.283^2=53.042$, $d_2^2 = 6.583^2 = 43.336$, $d_3^2=(-15.817)^2 = 250.177$, $d_4^2 = 3.683^2 = 13.565$, $d_5^2=(-12.317)^2 = 151.718$, $d_6^2 = 10.583^2 = 111.004$
The sum of squared differences $\sum d_i^2=53.042 + 43.336+250.177+13.565+151.718+111.004 = 622.842$
$s=\sqrt{\frac{\sum d_i^2}{n - 1}}=\sqrt{\frac{622.842}{5}}\approx\sqrt{124.568}\approx11.161$

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 $\mu = 61.5$, $n = 6$, $\bar{x}\approx89.117$, $s\approx11.161$
$t=\frac{89.117 - 61.5}{11.161/\sqrt{6}}=\frac{27.617}{11.161/2.449}\approx\frac{27.617}{4.56}{}\approx6.056$

Step4: Calculate the p - value

The degrees of freedom is $df=n - 1=6 - 1 = 5$. Since this is a right - tailed test, we find the p - value as $P(T>6.056)$ with $df = 5$. Using a t - distribution table or a calculator (e.g., in R: 1 - pt(6.056,5)), the p - value is approximately $0.001$

Step5: Make a decision

Since the significance level $\alpha = 0.10$ and the p - value ($0.001$) is less than $\alpha$, we reject the null hypothesis.

Answer:

test statistic: $6.056$
P - value: $0.001$
The P - value is less than (or equal to) $\alpha$
This test statistic leads to a decision to reject the null
The final conclusion is that the sample data support the claim that the population mean is greater than $61.5$