Question

Weight loss: In a study to determine whether counseling could help people lose weight, a sample of people experienced a group-based behavioral intervention, which involved weekly meetings with a trained interventionist for a period of six months. The following data are the numbers of pounds lost for 14 people, based on means and standard deviations given in the article. Assume the population is approximately normal. Perform a hypothesis test to determine whether the mean weight loss is greater than 12 pounds. Use the a=0.01 level of significance and the P-value method with the TI-84 Plus calculator. 18.6 25.2 4.1 20.4 17.3 9.0 19.8 13.7 15.5 17.6 34.0 29.9 8.7 31.7 Send data to Excel Part: 0 / 4 Part 1 of 4 (a) State the appropriate null and alternate hypotheses. HO: H: This hypothesis test is a (Choose one) test.

B) compute the P value

C ) determine whether to reject H0 use the a=0.01 level of significance

d) state a conclusion ( is or is not ) enough evidence to conclude greater than 12 pounds

Answer 1

$Since we know that \\Mean(\bar{x}) = \frac{\sum_{i=1}^n x_i}{n} \\\text{Where n is the number of data points} \\\text{Now} \\\sum_{i=1}^n x_i = 265.5 \\\text{and n = 14} \\\text{This implies that} \\Mean(\bar{x}) = \frac{265.5}{14} \\Mean(\bar{x}) = 18.9643 \\Variance(s^2) = \frac{(\sum{x_i - \bar{x}})^2}{n-1} \\\left(\sum{x_i - \bar{x}}\right)^2 = 1019.5723 \\n = 14 \\Variance(s^2) = \frac{1019.5723}{13} \\Variance(s^2) = 78.4286 \\\text{Standard Deviation(s) = }\sqrt{Variance} \\\text{Standard Deviation(s) = }8.856$

$\\\mu_0 = 12 \\n = 14 \\s = 8.856 \\\bar{X} = 18.9643 \\\alpha = 0.01 \\\text{Null Hypothesis }--> H_0: \mu \le 12 \\\text{Alternate Hypothesis }--> H_1: \mu > 12$

This is a right-tailed test because the alternative hypothesis is formulated to detect claim if difference of data is more than 0.

Now, the value of test static can be found out by following formula:

$\\t_0 = \frac{\bar{X} - \mu_0}{s/\sqrt{n}} \\t_0 = \frac{18.9643- 12.0}{8.856/\sqrt{14}} \\t_0 = 2.9424 \\\\\text{Using Excel's function }=T.DIST.RT(t_0,n-1)\text{, the P-value for }t_0 = 2.9424\text{ in an upper-tailed t-test with 13 degrees of freedom can be computed as P = }P(T_{13}>2.9424)=T.DIST.RT(2.9424,13)=0.005718827547082403. \\\\\text{Since P = 0.005718827547082403 < 0.01, we reject the null hypothesis }H_0:\mu= 12.0\text{ in favor of the alternative hypothesis }H_1:\mu>12.0\text{ at }\alpha = 0.01. \\\\\text{Critical value = }t_{\alpha, n-1} = t_{0.01, 13} = 2.6503 \\\\\text{Rejection Region: }t_0 > t_{\alpha, n-1} \\\\\text{Since }t_0 = 2.9424>2.6503=t_{0.01},\text{ we reject the null hypothesis }H_0:\mu=12.0\text{ in favor of the alternative hypothesis }H_1:\mu >12.0\text{ at }\alpha = 0.01.$

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