Question

# A food manufacturer claims that eating its new cereal as part of a daily diet lowers...

A food manufacturer claims that eating its new cereal as part of a daily diet lowers total blood cholesterol levels. The table shows the total blood cholesterol levels​ (in milligrams per deciliter of​ blood) of seven patients before eating the cereal and after one year of eating the cereal as part of their diets. Use technology to test the mean difference. Assume the samples are random and​ dependent, and the population is normally distributed. At alphaequals0.05​, can you conclude that the new cereal lowers total blood cholesterol​ levels?

Before   After
200   197
220   219
235   240
245   242
245   240
270   268
225   224

1. Let the blood cholesterol level after eating the cereal be population 2. Identify the null and alternative​ hypotheses, where mu Subscript dequalsmu 1 minus mu 2. Choose the correct answer below. A. Upper H 0​: mu Subscript dgreater than or equals0 Upper H Subscript Upper A​: mu Subscript dless than0 B. Upper H 0​: mu Subscript dless than or equals0 Upper H Subscript Upper A​: mu Subscript dgreater than0 C. Upper H 0​: mu Subscript dnot equals0 Upper H Subscript Upper A​: mu Subscript dequals0 D. Upper H 0​: mu Subscript dequals0 Upper H Subscript Upper A​: mu Subscript dnot equals0 Calculate the standardized test statistic. tequals nothing ​(Round to three decimal places as​ needed.) Calculate the​ P-value. ​P-valueequals nothing ​(Round to four decimal places as​ needed.) State the conclusion. ▼ Reject Fail to reject Upper H 0. There ▼ is not is sufficient evidence to support the claim that the new cereal lowers total blood cholesterol levels. Click to select your answer(s).

First, we need to estimate the mean and standard deviation for before and after

Before:

u1 = 238.5714

s1 = 23.0424

n1 = 7

After :

u2 = 232.8571

S2 = 22.2743

n2 = 7

Null hypothesis : Ho : d<0

Alternate hypothesis : Ha : d > 0

Test statistics = (u1-u2)/standard error

Standard error = √{(s1^2/n1)+(s2^2/n2)}

Test statistics = 0.472

Degrees of freedom is given by smaller of n1-1, n2-1

Df = 6

For df 6 and test statistics of 0.472

From t distribution

P-value is = 0.326802

As the obtained p-value is greater than the given significance level of 0.05

We fail to reject the null hypothesis Ho so we do not have enough evidence to support the claim

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