Question

Hypothesis Testing_02 (when is unknown) 9 Fifteen adult males between the ages of 35 and 50 participated in a study to evaluate the elec exercise on blood cholesterol levels. The total cholesterol was measured in each subject initially months after participating in an aerobic exercise program and switching to a low fat diet. The date the table below: in a study to evaluate the effect of diet and measured in each subject initially and then three tching to a low fat diet. The data are shown in Subject 229 240 240 WEWN Blood Cholesterol Level Before After 265 231 258 227 295 251 238 245 287 314 260 279 239 247 283 240 246 218 219 238 225 247 226 233 a. You need to test the claim that low-fat diet and aerobic exercise are of value in producing a mean reduction in blood cholesterol levels. Write down the null and alternative hypotheses to be tested. Clearly define all the terms. b. What is the type of statistical test that that should be used? Explain. C. Test the hypotheses at a = 0.05, using the critical value approach. What is your decision on null hypothesis(Ho)? Interpret the test results. d. What is the P Value for the test? Test the hypotheses using the P Value. Do you get the same answer as part (c)? e. Construct a 95% confidence interval for the mean difference of Blood Cholesterol levels. Test the hypotheses using the confidence interval approach. Do you get the same answer as parts (c) and (d)?

Answer 1

a). hypothesis:-

$H_{0}:\mu_{d}=0$

$H_{1}:\mu_{d}>0$

where, $\mu_{d}$ is the difference between mean blood cholesterol levels between before and  after participating in an aerobic exercise program and switching to a low fat diet.

our claim is the alternative hypothesis.

b).here,we will do , paired t test

because blood cholesterol levels were measured initially and  after participating in an aerobic exercise program and switching to a low fat diet.so, the data is dependent..so we will perform paired t test.

c). necessary calculation table:-

before($x_{i}$)	after($y_{i}$)	$d_{i}=x_{i}-y_{i}$	$(d_{i}-\bar{d})^2$
265	229	36	83.41717
240	231	9	319.219
258	227	31	17.08417
295	240	55	791.4826
251	238	13	192.2854
245	241	4	522.886
287	234	53	682.9494
314	256	58	969.2824
260	247	13	192.2854
279	239	40	172.4836
283	246	37	102.6838
240	218	22	23.68477
238	219	19	61.88497
225	226	-1	776.553
247	233	14	165.552
		sum=403	sum= 5073.7

$\bar{d}=\frac{\sum d_{i}}{n}=\frac{403}{15}=26.8667$

$s_{d}=\sqrt{\frac{\sum (d_{i}-\bar{d})^2}{n-1}}=\sqrt{\frac{5073.7}{15-1}}=19.037$

the test statistic be:-

$t=\frac{\bar{d}}{s_{d}/\sqrt{n}}=\frac{26.8667}{19.037/\sqrt{15}}=5.466$

degrees of freedom = (n-1) = (15-1) = 14

t critical value for df = 14, alpha= 0.05, one tailed test be:-

$t^*=1.761$ [ from t table]

decision:-

$t=5.466>t^*=1.761$

so, we reject the null hypothesis.

conclusion:-

there is sufficient evidence to support the claim that low-fat diet and aerobic exercise are of value in producing a mean reduction in blood cholesterol levels at 0.05 level of significance.

d). p value = 0.0000

[ using calculator, for t = 5.466, df= 14, right tailed test ]

decision based on p value:-

p value = 0.0000 <0.05,

so, we reject the null hypothesis.

YES,i have obtained the same answer as in part c.

i.e, there is sufficient evidence to support the claim that low-fat diet and aerobic exercise are of value in producing a mean reduction in blood cholesterol levels at 0.05 level of significance.

e). the 95 % confidence interval of difference in mean be:-

$=(\bar{d}-[t^**\frac{s_{d}}{\sqrt{n}}],\infty )$ [ as, it is a right tailed test ]

$=(26.8667-(1.761*\frac{19.037}{\sqrt{15}}),\infty )$

$=(18.211,\infty )$

decision:-

the hypothesized value (0) is not contained in the interval,

we reject the null hypothesis.

hence again we conclude that,

there is sufficient evidence to support the claim that low-fat diet and aerobic exercise are of value in producing a mean reduction in blood cholesterol levels at 0.05 level of significance.

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