Respuesta :
Answer:
a) [tex]t=\frac{4.6-5}{\frac{2.2}{\sqrt{20}}}=-0.813[/tex]
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=20-1=19[/tex]
[tex]p_v =P(t_{(19)}<-0.813)=0.213[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is lower than 5 years at 5% of signficance.
b) [tex]4.6-2.09\frac{2.2}{\sqrt{20}}=3.572[/tex]
[tex]4.6+2.09\frac{2.2}{\sqrt{20}}=5.628[/tex]
So on this case the 95% confidence interval would be given by (3.572;5.628)
c) Yes from the hypothesis test we fail to reject the null hypothesis that the true mean is equal or higher to 5 and from the confidence interval the limits contains the value 5, so then the two procedures are showing the same result FAIL to reject the null hypothesis.
Step-by-step explanation:
Data given and notation
[tex]\bar X=4.6[/tex] represent the sample mean
[tex]s=2.2[/tex] represent the sample standard deviation
[tex]n=20[/tex] sample size
[tex]\mu_o =5[/tex] represent the value that we want to test
[tex]\alpha=0.05[/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)
Part a
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is at least 5 years, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 5[/tex]
Alternative hypothesis:[tex]\mu < 5[/tex]
If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{4.6-5}{\frac{2.2}{\sqrt{20}}}=-0.813[/tex]
P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=20-1=19[/tex]
Since is a one sided test the p value would be:
[tex]p_v =P(t_{(19)}<-0.813)=0.213[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is lower than 5 years at 5% of signficance.
Part b
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=20-1=19[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,19)".And we see that [tex]t_{\alpha/2}=2.09[/tex]
Now we have everything in order to replace into formula (1):
[tex]4.6-2.09\frac{2.2}{\sqrt{20}}=3.572[/tex]
[tex]4.6+2.09\frac{2.2}{\sqrt{20}}=5.628[/tex]
So on this case the 95% confidence interval would be given by (3.572;5.628)
And as we can see the confidence interval contains the value 5.
Part c
Yes from the hypothesis test we fail to reject the null hypothesis that the true mean is equal or higher to 5 and from the confidence interval the limits contains the value 5, so then the two procedures are showing the same result FAIL to reject the null hypothesis.
Using the t-distribution, it is found that:
a) Since the p-value of the test is 0.214 > 0.05, there is not significant evidence to conclude that the mean is less than 5 years, hence, Georgianna's claim cannot be rejected.
b) The 95% confidence interval for the number of years students in this city take piano lessons is (3.57, 5.63). The upper bound of the interval is higher than 5, hence it cannot be concluded that the mean is less than 5 years and Georgianna's claim cannot be rejected.
c) In both cases, using the p-value and the confidence interval, her claim cannot be reject, hence, the results agree.
Item a:
At the null hypothesis, we test if the mean is of at least 5 years, that is:
[tex]H_0: \mu \geq 5[/tex]
At the alternative hypothesis, we test if the mean is of less than 5 years, that is:
[tex]H_1: \mu < 5[/tex]
We have the standard deviation for the sample, thus, the t-distribution is used. The test statistic is given by:
[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]
The parameters are:
[tex]\overline{x}[/tex] is the sample mean.
[tex]\mu[/tex] is the value tested at the null hypothesis.
s is the standard deviation of the sample.
n is the sample size.
For this problem, the values of the parameters are: [tex]\overline{x} = 4.6, \mu = 5, s = 2.2, n = 20[/tex]
Hence, the value of the test statistic is:
[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{4.6 - 5}{\frac{2.2}{\sqrt{20}}}[/tex]
[tex]t = -0.81[/tex]
The p-value of the test is found using a left-tailed test, as we are testing if the mean is less than a value, with t = -0.81 and 20 - 1 = 19 df.
Using a t-distribution calculator, this p-value is of 0.214.
Since the p-value of the test is 0.214 > 0.05, there is not significant evidence to conclude that the mean is less than 5 years, hence, Georgianna's claim cannot be rejected.
Item b:
The interval is given by:
[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]
The critical value, using a t-distribution calculator, for a two-tailed 95% confidence interval, with 19 df, is t = 2.093.
Hence:
[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 4.6 - 2.093\frac{2.2}{\sqrt{20}} = 3.57[/tex]
[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 4.6 + 2.093\frac{2.2}{\sqrt{20}} = 5.63[/tex]
The 95% confidence interval for the number of years students in this city take piano lessons is (3.57, 5.63). The upper bound of the interval is higher than 5, hence it cannot be concluded that the mean is less than 5 years and Georgianna's claim cannot be rejected.
Item c:
In both cases, using the p-value and the confidence interval, her claim cannot be reject, hence, the results agree.
A similar problem is given at https://brainly.com/question/12491757
