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user100 [1]
3 years ago
5

In their advertisements, the manufacturers of a diet pill claims that taken daily, their pill will produce an average weight los

s of 10 pounds in one month, with standard deviation 3 pounds. In order to determine if this is a valid claim, an independent testing agency selects 36 people to take the pill daily for a month and found the mean weight loss for this group was 9 pounds.
1. Does this indicate at a 5% level that the mean weight loss is less than 10 pounds?
Mathematics
2 answers:
Lina20 [59]3 years ago
8 0

Answer:

We conclude that the mean weight loss is less than 10 pounds.

Step-by-step explanation:

We are given that the manufacturers of a diet pill claims that taken daily, their pill will produce an average weight loss of 10 pounds in one month, with standard deviation 3 pounds.

In order to determine if this is a valid claim, an independent testing agency selects 36 people to take the pill daily for a month and found the mean weight loss for this group was 9 pounds.

Let  = mean weight loss.

So, Null Hypothesis,  :  10 pounds     {means that the mean weight loss is more than or equal to 10 pounds}

Alternate Hypothesis,  :  < 10 pounds     {means that the mean weight loss is less than 10 pounds}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

                      T.S. =    ~ N(0,1)

where,  = sample mean weight loss = 9 pounds

            = population standard deviation = 10 pounds

           n = sample of people = 36

Step-by-step explanation:

maxonik [38]3 years ago
3 0

Answer:

We conclude that the mean weight loss is less than 10 pounds.

Step-by-step explanation:

We are given that the manufacturers of a diet pill claims that taken daily, their pill will produce an average weight loss of 10 pounds in one month, with standard deviation 3 pounds.

In order to determine if this is a valid claim, an independent testing agency selects 36 people to take the pill daily for a month and found the mean weight loss for this group was 9 pounds.

<u><em>Let </em></u>\mu<u><em> = mean weight loss.</em></u>

So, Null Hypothesis, H_0 : \mu \geq 10 pounds     {means that the mean weight loss is more than or equal to 10 pounds}

Alternate Hypothesis, H_A : \mu < 10 pounds     {means that the mean weight loss is less than 10 pounds}

The test statistics that would be used here <u>One-sample z test statistics</u> as we know about the population standard deviation;

                       T.S. =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean weight loss = 9 pounds

            \sigma = population standard deviation = 10 pounds

            n = sample of people = 36

So, <u><em>test statistics</em></u>  =  \frac{9-10}{\frac{3}{\sqrt{36} } }

                               =  -2

The value of z test statistics is -2.

Now, at 5% significance level the z table gives critical value of -1.645 for left-tailed test. Since our test statistics is less than the critical value of z as -2 < -1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that the mean weight loss is less than 10 pounds.

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