Question

A recent study found that 51 children who watched a commercial for Walker Crisps (potato chips) featuring a long-standing sports celebrity endorser ate a mean of 36 grams of Walker Crisps as compared to a mean of 25 g of Walker Crisps for 41 children who watched a commercial for an alternative food snack. Suppose that the sample standard deviation for the children who watched the sports celebrity–endorsed Walker Crisps commercial was 21.4 g and the sample standard deviation for the children who watched the alternative food snack commercial was 12.8 g. Assuming the population variances are not equal and alpha-05, is there any evidence that the mean amount of Walker Crisps eaten was significantly higher for the children who watched the sports celebrity- endorsed Walker Crisps commercial?
1. What is the claim from the question? What are Null and Alternative Hypothesis for this problem?
2. What kind of test do you want to use?
3. Calculate Test Statistics.
4. Find P-value.
5. What is the conclusion that you could make?

Answer 1

Answer:

1

The claim is that the mean amount of Walker Crisps eaten was significantly higher for the children who watched the sports celebrity- endorsed Walker Crisps commercial

2

The kind of test to use is a t -test because a t -test is used to check if there is a difference between means of a population

3

$t = 3.054$

4

The p-value  is   $p-value = P(Z > 3.054) = 0.0011291$

5

The conclusion is  

There is sufficient evidence to conclude that the mean amount of Walker Crisps eaten was significantly higher for the children who watched the sports celebrity- endorsed Walker Crisps commercial

The test statistics is  

Step-by-step explanation:

From the question we are told that

   The first sample size is  $n_1 = 51$

    The first sample  mean is $\mu_1 = 36$

    The second sample size is  $n_2 = 41$

    The second sample  size is  $\mu_2 = 25$

     The first standard deviation is  $\sigma _1 = 21.4 \ g$

    The second standard deviation is  $\sigma _2 = 12.8 \ g$

  The  level of significance is  $\alpha = 0.05$

The  null hypothesis is  $H_o : \mu_1 = \mu_ 2$

The  alternative hypothesis is  $H_a : \mu_1 > \mu_2$

Generally the test statistics is mathematically represented as

    $t = \frac{\= x_1 - \= x_2}{ \sqrt{ \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} } }$

=>   $t = \frac{ 36 - 25}{ \sqrt{ \frac{ 21.4^2}{51} + \frac{ 12.8^2}{41} } }$

=> $t = 3.054$

The  p-value is mathematically represented as

     $p-value = P(Z > 3.054)$

Generally from the z table  

             $P(Z > 3.054) = 0.0011291$

=>   $p-value = P(Z > 3.054) = 0.0011291$

From the values obtained  we see that $p-value < \alpha$ so  the null hypothesis is rejected

Thus the conclusion is  

  There is sufficient evidence to conclude that the mean amount of Walker Crisps eaten was significantly higher for the children who watched the sports celebrity- endorsed Walker Crisps commercial