Respuesta :
Question:
Company 2007 Score 2008 Score
Rite Aid 73 76
Expedia 75 77
J.C. Penney 77 78
a. For Rite Aid, is the increase in the satisfaction score from 2007 to 2008 statistically significant? Use α= .05. What can you conclude?
b. Can you conclude that the 2008 score for Rite Aid is above the national average of 75.7? Use α= .05.
c. For Expedia, is the increase from 2007 to 2008 statistically significant? Use α= .05.
d. When conducting a hypothesis test with the values given for the standard deviation,
sample size, and α, how large must the increase from 2007 to 2008 be for it to be statistically significant?
e. Use the result of part (d) to state whether the increase for J.C. Penney from 2007 to 2008 is statistically significant.
Answer:
a. There is sufficient statistical evidence to suggest that the increase in satisfaction score for Rite Aid from 2007 to 2008 is statistically significant
b. There is sufficient statistical evidence to suggest that the 2008 Rite Aid score, is above the national average of 75.7
c. The statistical evidence support the claim of a significant increase from 2007 to 2008
d. 1.802 and above is significant
e. The increase of J. C. Penney from 2007 is not statistically significant.
Step-by-step explanation:
Here we have
n = 60
σ = 6
μ₁ = 73
μ₂ = 76
We put H₀ : μ₁ ≥ μ₂ and
Hₐ : μ₁ < μ₂
From which we have;
[tex]z=\frac{(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2} }{n_{1}}+\frac{\sigma _{2}^{2}}{n_{2}}}} = \frac{(\mu_{1}-\mu_{2})}{\sqrt{\frac{2\sigma_{}^{2} }{n_{}}}}}[/tex]
Plugging in the values we have
[tex]z = \frac{(73-76)}{\sqrt{\frac{2\times 6^{2} }{60_{}}}}} = -2.7386[/tex]
The probability, P from z function computation gives;
P(Z < -2.7386) = 0.0031
Where we have P < α, we reject the null hypothesis meaning that there is sufficient statistical evidence to suggest that the increase in satisfaction score for Rite Aid from 2007 to 2008 is statistically significant
b. To test here, we have
H₀ : μ ≤ 75.7
Hₐ : μ > 75.7
The test statistic is given as follows;
[tex]z=\frac{\bar{x}-\mu }{\frac{\sigma }{\sqrt{n}}} = \frac{76-75.7 }{\frac{6 }{\sqrt{60}}} = 0.3873[/tex]
Therefore, we have the probability, P given as the value for the function at z = 0.3873 that is we have;
P = P(Z > 0.3873) = P(Z < -0.3873) = 0.3493
Therefore, since P > α which is 0.05, we fail to reject the null hypothesis, that is there is sufficient statistical evidence to suggest that the 2008 Rite Aid score, is above the national average of 75.7
c. Here we put
Null hypothesis H₀ : μ₁ ≥ μ₂
Alternative hypothesis Hₐ : μ₁ < μ₂
The test statistic is given by the following equation;
[tex]z=\frac{(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2} }{n_{1}}+\frac{\sigma _{2}^{2}}{n_{2}}}} = \frac{(\mu_{1}-\mu_{2})}{\sqrt{\frac{2\sigma_{}^{2} }{n_{}}}}}[/tex]
Plugging in the values we have
[tex]z = \frac{(75-77)}{\sqrt{\frac{2\times 6^{2} }{60_{}}}}} = -1.8257[/tex]
The probability, P from z function computation gives;
P(Z < -1.8257) = 0.03394
The statistical evidence support the claim of a significant increase
d. For statistical significance at 0.05 significant level, we have z = -1.644854
Therefore, from;
[tex]z=\frac{(\bar{x_{1}}-\bar{x_{2}})-(\mu_{1}-\mu _{2} )}{\sqrt{\frac{\sigma_{1}^{2} }{n_{1}}-\frac{\sigma _{2}^{2}}{n_{2}}}}[/tex]. we have;
[tex]z \times \sqrt{\frac{\sigma_{1}^{2} }{n_{1}}+\frac{\sigma _{2}^{2}}{n_{2}}} + (\mu_{1}-\mu _{2} )}{} ={(\bar{x_{1}}-\bar{x_{2}})[/tex]
Which gives
[tex]{(\bar{x_{1}}-\bar{x_{2}}) = z \times \sqrt{\frac{2\sigma_{}^{2} }{n_{}}}} + (\mu_{1}-\mu _{2} )}{} = -1.644854 \times \sqrt{\frac{2\times 6_{}^{2} }{60_{}}}} + 0 = -1.802[/tex]
Therefore an increase of 1.802 and above is significant
e. Based on the result of part d. we have for J.C. Penney from 2007 to 2008 an increase of 1 which is less than 1.802 at 5% significant level, is not significant.
