Respuesta :
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
[tex]t = 3.054[/tex]
4
The p-value is [tex]p-value = P(Z > 3.054) = 0.0011291[/tex]
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 [tex]n_1 = 51[/tex]
The first sample mean is [tex]\mu_1 = 36[/tex]
The second sample size is [tex]n_2 = 41[/tex]
The second sample size is [tex]\mu_2 = 25[/tex]
The first standard deviation is [tex]\sigma _1 = 21.4 \ g[/tex]
The second standard deviation is [tex]\sigma _2 = 12.8 \ g[/tex]
The level of significance is [tex]\alpha = 0.05[/tex]
The null hypothesis is [tex]H_o : \mu_1 = \mu_ 2[/tex]
The alternative hypothesis is [tex]H_a : \mu_1 > \mu_2[/tex]
Generally the test statistics is mathematically represented as
[tex]t = \frac{\= x_1 - \= x_2}{ \sqrt{ \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} } }[/tex]
=> [tex]t = \frac{ 36 - 25}{ \sqrt{ \frac{ 21.4^2}{51} + \frac{ 12.8^2}{41} } }[/tex]
=> [tex]t = 3.054[/tex]
The p-value is mathematically represented as
[tex]p-value = P(Z > 3.054)[/tex]
Generally from the z table
[tex]P(Z > 3.054) = 0.0011291[/tex]
=> [tex]p-value = P(Z > 3.054) = 0.0011291[/tex]
From the values obtained we see that [tex]p-value < \alpha[/tex] 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
The value of the difference in the mean number of chips eaten can be used to test the hypothesis.
The correct responses are;
- 1. The claim is that children that watched the given commercial ate more Walker Crisps potato chips than others.
- The null hypothesis is [tex]\underline {H_0; \ \overline x_1 = \overline x_2}[/tex] , the alternative hypothesis is, [tex]\underline{H_a; \ \overline x_1 > \overline x_2}[/tex]
- 2. The kind of test to use is a t-test
- 3. The test statistic is approximately 3.054
- 4. The p-value is; 0.001 < p-value < 0.0025
- 5. Reject the null hypothesis, there is significant statistical evidence to suggest that the children that watched the commercial ate more Walter Crisps potato chips than children that watched the commercial for an alternative snack.
Reasons:
The given parameters are;
The amount of Walker Crisps potato chips children ate, based on commercial watched.
Celebrity endorsed food snack commercial mean, [tex]\overline x_1[/tex] = 36 g
Sample size that watched the celebrity endorsed commercial, n₁ = 51
Alternative food snack commercial mean, [tex]\overline x_2[/tex] = 25 g
Sample size that watched the other commercial = 41
Standard deviation, SD, of the samples;
Sample that watched the celebrity endorsed commercial, SD₁ = 21.4 g
Sample that watched the other commercial, SD₂ = 12.8 g
Reasons:
1. The claim from the question is; The mean amount of Walker Crisp eaten was significantly higher for the children who watched the sports celebrity - endorsed Walker Crisps commercial.
The null hypothesis is, H₀; [tex]\overline x_1 = \overline x_2[/tex] (There is no difference in the mean)
The alternative hypothesis is, Hₐ; [tex]\overline x_1[/tex] > [tex]\overline x_2[/tex] (The mean of the amount of Walker Crisps eaten is significantly higher for the children that watched the sports celebrity-endorsed commercial)
2. Given that the aim of the study is to determine if the mean of the first
sample is higher than the mean of the second sample, the kind of test to
be performed is the test of the difference in the mean of the two samples
based on specified hypothesis, (one sample is higher), which is done using
a t-test, given that the population mean is not specified (unknown).
The kind of test to use is a t-test
3. The test statistic is given by Welch's Test for unequal variance as follows;
[tex]\displaystyle t = \mathbf{\frac{\overline x_1 - \overline x_2}{\sqrt{\dfrac{s_1^2}{n_1} +\dfrac{s_2^2}{n_2} } }}[/tex]
Which gives;
[tex]\displaystyle t = \frac{36- 25}{\sqrt{\dfrac{21.4^2}{51} +\dfrac{12.8^2}{41} } } = \frac{77500}{25379} \approx 3.054[/tex]
The test statistic, t ≈ 3.054
4. From the above test statistic and by finding, degrees of freedom, df as follows, the p-value can be found.
[tex]\displaystyle df = \mathbf{\frac{\left(\dfrac{s_1^2}{n_1} +\dfrac{s_2^2 \righ}{n_2} \right)^2}{\dfrac{1}{n_1 - 1} \cdot \left( \dfrac{s_1^2}{n_1} \right)^2 + \dfrac{1}{n_2 - 1} \cdot \left( \dfrac{s_2^2}{n_2} \right)^2}}[/tex]
Which gives;
[tex]\displaystyle df = \frac{\left(\dfrac{21.4^2}{51} +\dfrac{12.8^2 \righ}{41} \right)^2}{\dfrac{1}{51 - 1} \cdot \left( \dfrac{21.4^2}{51} \right)^2 + \dfrac{1}{41 - 1} \cdot \left( \dfrac{12.8^2}{41} \right)^2} \approx \mathbf{83.69}[/tex]
From the t-test table, we have the p-value of the result as follows;
The p-value is; 0.001 < P(t > 3.054) < 0.0025.
5. The conclusion that can be made is; Given that the p-value of the test
statistic is less than the given alpha value of 0.05, we reject the null
hypothesis, given that there is significant statistical evidence to show that
the mean amount of Walker Crisp eaten was significantly higher for the
children that watch the celebrity-endorsed Walker Crisps commercial.
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