Question

A diet center claims that it has the most effective weight loss program in the region. Its advertisements say, "Participants in our program lose more than 4 pounds within a month." Six clients of this program are weighed on the first day of the diet and then one month later. Let the difference be defined as Weight on First Day of Diet minus Weight One Month Later. (You may find it useful to reference the appropriate table: z table or t table) Weight on First Weight One Day of Diet Month Later Client 167 209 168 206 152 141 154 204 162 183 156 149 Xc Let the difference be defined as Before After a. Specify the null and alternative hypotheses that test the diet center's claim. b. Assuming that weight loss is normally distributed, calculate the value of the test statistic. (Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) Test statistic

Answer 1

a)

Null and alternative hypothesis :

$H_0 : \mu_D \leq 4$

$H_a : \mu_D > 4$ ( Claim )

One tailed test ( Right tailed )

b)

Test statistic :

$t = \frac{\bar{x}_D - \mu _D}{\frac{s_D}{\sqrt{n}}}$

Where , $\bar{x}_D$ is sample mean of differences.

$s_D$ is sample standard deviation of differences.

n is sample size.

$\bar{x}_D$ = 5.8333 , n = 6 , $s_D$ = 11.2679 , $\mu _D$ = 4

So, test statistics is ,

$t = \frac{5.8333-4}{\frac{11.2679}{\sqrt{6}}} = 0.3985$

Test Statistic = 0.40 ( Rounded to 2 decimal places )

P-value -

P-value for this right tailed test is given by,

P-value = P( t > test statistic) = P( t > 0.3985 )

Using Excel,   =TDIST( t , df , tails ) , df = n - 1 =6 - 1 = 5 , tails = 1

P-value = P( t > 0.3985 ) = TDIST(0.3985 , 5 , 1 ) = 0.3534

P-value = 0.35

Decision about null hypothesis -

Let significance level = $\alpha =0.05$

P-value ( 0.35 ) is greater than $\alpha =0.05$

So, fail to reject null hypothesis.

Conclusion -

There is not sufficient evidence to conclude that participants lose more than 4 pound within a month.