Respuesta :
Answer:
[tex]z=\frac{85-88}{\sqrt{\frac{7^2}{30}+\frac{5^2}{30}}}}=-1.91[/tex]
[tex]p_v =2*P(Z<-1.91)=0.0561[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can't conclude that the two means are significantly different at 1% of significance.
Step-by-step explanation:
Data given and notation
[tex]\bar X_{A}=85[/tex] represent the mean for the sample of 2016-2017
[tex]\bar X_{B}=88[/tex] represent the mean for the sample of the last year
[tex]\sigma_{A}=7[/tex] represent the population standard deviation for the sample of 2016-2017
[tex]\sigma_{B}=5[/tex] represent the population standard deviation for the sample of last year
[tex]n_{A}=30[/tex] sample size selected for the sample of 2016-2017
[tex]n_{B}=30[/tex] sample size selected for the last year
[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the two means are equal or no, the system of hypothesis would be:
Null hypothesis:[tex]\mu_{A} = \mu_{B}[/tex]
Alternative hypothesis:[tex]\mu_{A} \neq \mu_{B}[/tex]
Since we know the population deviations, for this case is better apply a z test to compare means, and the statistic is given by:
[tex]z=\frac{\bar X_{A}-\bar X_{B}}{\sqrt{\frac{\sigma^2_{A}}{n_{A}}+\frac{\sigma^2_{B}}{n_{B}}}}[/tex] (1)
z-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]z=\frac{85-88}{\sqrt{\frac{7^2}{30}+\frac{5^2}{30}}}}=-1.91[/tex]
P-value
Since is a two tailed test the p value would be:
[tex]p_v =2*P(Z<-1.91)=0.0561[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can't conclude that the two means are significantly different at 1% of significance.
We cannot conclude that the two means are significantly different at 0.01 of significance.
How to determine the conclusion
Sample size of both classes, n = 30
Sample mean of 2016-2017, [tex]\bar x_a = 85\%[/tex]
Sample mean of current year, [tex]\bar x_a = 88\%[/tex]
Sample standard deviation of 2016-2017, [tex]\sigma_b = 7\%[/tex]
Sample standard deviation of current year, [tex]\sigma_b = 5\%[/tex]
Population mean, [tex]\mu = 13[/tex]
Sample standard deviation, σ = 0.4
In this case we want to determine if the two means are equal or not
So, the hypotheses are:
Null [tex]H_o : \mu_a = \mu_b[/tex]
Alternate, [tex]H_a : \mu_a \ne \mu_b[/tex]
The test statistic is then calculated as:
[tex]t = \frac{\bar x_a - \bar x_b}{\sqrt{\sigma_a^2/n + \sigma_b^2/n}}[/tex]
So, we have:
[tex]t = \frac{85\%- 88\%}{\sqrt{7\%^2/30 + 5\%^2/30}}[/tex]
[tex]t = -1.91[/tex]
Calculate the p value
For a two-tailed test, we have:
[tex]p = 2 * P(t < -1.91)[/tex]
Using the p table of probabilities, we have:
[tex]p = 0.056[/tex]
The p value is greater than the level of significance.
This means that we fail to reject the null hypothesis
Hence, we cannot conclude that the two means are significantly different at 0.01 of significance.
Read more about test statistic at:
brainly.com/question/16628265
