Question

Assembly Time: In a sample of 45 adults, the mean assembly time for a child's swing set was 1.78 hours with a standard deviation of 0.80 hours. The makers of the swing set daim the average assembly time is less than 2 hours. Test their daim at the 0.05 significance level. (a) What type of test is this? This is a left-tailed test. This is a two-tailed test. This is a right-tailed test. (b) What is the test statistic Round your answer to 2 decimal places. t; -1.84 (c) Use software to get the P-value of the test statistic. Round to 4 decimal places. P-value - 0718 x (d) What is the condusion regarding the null hypothesis? reject Ho • fail to reject Ho (e) Choose the appropriate concluding statement. The data supports the claim that the mean assembly time is less than 2 hours There is not enough data to support the claim that the mean assembly time is less than 2 hours. We reject the daim that the mean assembly time is less than 2 hours We have proven that that the mean assembly time is less than 2 hours.

Answer 1

$\small \mu_0 = 2$
n = 45
$\small \\s = 0.8 \\\bar{X} = 1.78 \\\alpha = 0.05$
The test hypothesis is
$\small \\Null\;Hypothesis --> H_0: \mu \ge \mu_0 \\Alternate\;Hypothesis --> H_1: \mu < \mu_0$
a) This is a left tailed test
b) Now, the value of test static can be found out by following formula:
$\small \\t_0 = \frac{\bar{X} - \mu_0}{s/\sqrt{n}} \\t_0 = \frac{1.78- 2.0}{0.8/\sqrt{45}} \\t_0 = -1.84$
c) Using Excel's function =
T.DIST(to.n - 1.TRUE)
, the P-value for $\small t_0 = -1.8448$ in an power-tailed t-test with 44 degrees of freedom can be computed as $\small P = P(T_{44}<-1.8448)=T.DIST(-1.8448,44,TRUE)=0.0359$ .
d) Since P = 0.0359 < 0.05, we reject the null hypothesis $\small H_0:\mu= 2.0$ in favor of the alternative hypothesis $\small H_1:\mu>2.0\ at\ \alpha = 0.05.$
Since the sample size is n = 45, degrees of freedom on the t-test statistic are n-1 = 45-1 = 44
This implies that
$\small t_{\alpha, n-1} = t_{0.05, 44} = 1.68$
Since, the t distribution is symmetric about zero, so -t_{0.05,44}
Since $\small t_0 = -1.8448<-1.68=-t_{0.05}$ , we reject the null hypothesis $\small H_0:\mu= 2.0$ in favor of the alternative hypothesis $\small H_1:\mu>2.0\ at\ \alpha = 0.05.$ .

e) The data supports the claim that the mean assembly time is less than 2 hours.