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Answer :
Answer:
a) [tex]p_v =P(t_{(9)}<-2.679) =0.0126[/tex]
The p value is higher than the significance level given 0.01, so then we can conclude that we FAIL to reject the null hypothesis. And we can say that the true difference for Best Actresses is not significantly lower than the mean for Best Actors at 1% of significance.
b) The 99% confidence interval would be given by (-21.469;2.069)
c) We got the same conclusion as part a, sicne the confidence interval contains the value 0, we FAIL to reject the null hypothesis that the difference between the two
Step-by-step explanation:
Part a
Let put some notation
x=actor's age , y = actress's age
x: 58 41 36 36 34 33 48 37 37 43
y: 26 27 34 26 35 29 23 42 30 34
The system of hypothesis for this case are:
Null hypothesis: [tex]\mu_y- \mu_x \geq 0[/tex]
Alternative hypothesis: [tex]\mu_y -\mu_x <0[/tex]
The first step is calculate the difference [tex]d_i=y_i-x_i[/tex] and we obtain this:
d: -32, -14, -2, -10, 1, -4, -25, 5, -7, -9
The second step is calculate the mean difference
[tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}= -9.7[/tex]
The third step would be calculate the standard deviation for the differences, and we got:
[tex]s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =11.451[/tex]
The 4 step is calculate the statistic given by :
[tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-9.7 -0}{\frac{11.451}{\sqrt{10}}}=-2.679[/tex]
The next step is calculate the degrees of freedom given by:
[tex]df=n-1=10-1=9[/tex]
Now we can calculate the p value, since we have a left tailed test the p value is given by:
[tex]p_v =P(t_{(9)}<-2.679) =0.0126[/tex]
The p value is higher than the significance level given 0.01, so then we can conclude that we FAIL to reject the null hypothesis. And we can say that the true difference for Best Actresses is not significantly lower than the mean for Best Actors at 1% of significance.
Part b
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The confidence interval for the mean is given by the following formula:
[tex]\bar d \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,9)".And we see that [tex]t_{\alpha/2}=3.25[/tex]
Now we have everything in order to replace into formula (1):
[tex]-9.7-3.25\frac{11.451}{\sqrt{10}}=-21.469[/tex]
[tex]-9.7+3.25\frac{11.451}{\sqrt{10}}=2.069[/tex]
So on this case the 99% confidence interval would be given by (-21.469;2.069)
Part c
We got the same conclusion as part a, sicne the confidence interval contains the value 0, we FAIL to reject the null hypothesis that the difference between the two means is 0.
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Final answer:
The null hypothesis is that there's no difference in the mean ages of Best Actresses and Best Actors. The alternative hypothesis is that Best Actresses are generally younger than Best Actors. The confidence interval, calculated from the differences in the paired ages, supports either the null or alternative hypothesis based on whether it includes or excludes 0.
Explanation:
The null and alternative hypotheses for this hypothesis test are as follows:
Null hypothesis (H₀): μd = 0 (This suggests that there's no difference in the average age between Best Actresses and Best Actors.)
Alternative hypothesis (Hₐ): μd < 0 (This supports the claim that Best Actresses are generally younger than Best Actors.)
For part (b), to construct the confidence interval, you would first take all 10 differences (actress’s age minus the actor’s age) for each pair of ages. Then calculate the mean and standard deviation of these differences. After that, use the t-distribution to calculate the confidence interval with a 0.01 significance level.
If the confidence interval includes 0, it supports the null hypothesis that there's no significant difference in age. If the interval is entirely less than 0 (e.g. (-5, -1)), this indicates the mean difference is less than 0, supporting the alternative hypothesis that actresses are typically younger.
Learn more about Hypothesis Testing here:
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