Answer :
1. The calculated test statistic is approximately -0.326.
2. The p-value is about 0.373, so we fail to reject the null hypothesis.
3a) The hypothesis test involves a normal distribution and uses the t-distribution.
3b) The test uses the z-distribution for a normally distributed population with a known standard deviation.
4. The computed test statistic for the paired sample data is approximately -0.4862.
5. For a 99% confidence interval, the Z-score is approximately -2.576. Unfortunately, I can't interpret or draw missing parts of a graph without more specific details.
1. Given data:
Sample size (n) = 25
Sample mean (X) = 24.4 years
Sample standard deviation (s) = 9.2 years
Significance level (α) = 0.05
The null hypothesis (H₀) states that the mean age of the prison population is 25 years or more. The alternative hypothesis (H₁) states that the mean age is less than 25 years.
The test statistic for a one-sample t-test is given by:
t= X-μ₀/s/√n
Where:
X is the sample mean
μ₀ is the hypothesized population mean under the null hypothesis (25 years in this case)
s is the sample standard deviation
n is the sample size
Substituting the values:
t = 24.4-25/9.2/√25
Calculating the numerator and denominator separately:
t = -0.6/1.84
Now calculating the test statistic:
t≈−0.326
Rounded to 3 decimal places, the test statistic is approximately -0.326.
2. Since this is a one-tailed test (claiming that the mean age is less than 25), we need to find the p-value corresponding to the test statistic -0.326. You can use a t-distribution table, a calculator, or statistical software to find the p-value.
Using the t-distribution with degrees of freedom df = n-1 = 25-1 = 24, the p-value for t=−0.326 is approximately 0.373.
Since the p-value (0.373) is greater than the significance level (α = 0.05), we fail to reject the null hypothesis. There is not enough evidence to conclude that the mean age of the prison population is less than 25 years.
3. a. For a claim involving a normally distributed population with unknown mean and standard deviation, we use the t-distribution.
b. Similarly, for the second claim involving a normally distributed population with known standard deviation (σ = 30), we use the z-distribution.
4. Given paired sample data:
X:28,31,20,25,26,27,33,35
Y:26,27,26,25,29,32,33,34
The paired difference d = X - Y for each pair:
d:2,4,−6,0,−3,−5,0,1
Mean of the differences ([tex]\bar d[/tex]) = (2 + 4 - 6 + 0 - 3 - 5 + 0 + 1) / 8 = -1.125
Standard deviation of the differences (sd) = 3.940
Sample size (n) = 8
Hypothesized population mean difference (d₀) = 0
The test statistic for a paired t-test is given by:
t= d-d₀/sd/√n
Substituting the values:
t=-1.125-0/3.940/√8
t≈−0.4862
Rounded to four decimal places, the test statistic is approximately -0.4862.
5. For a 99% confidence interval, the critical z-score is the z-score corresponding to the middle 99% of the standard normal distribution. This can be found using a standard normal distribution table or calculator.
The area in the middle (between -Z and Z) for a 99% confidence interval is 0.99.
So, the area to the left (and to the right) is (1 - 0.99) / 2 = 0.005.
The Z-score for a cumulative area of 0.005 is approximately -2.576.
For a 99% confidence interval, the Z-score is -2.576.
To know more about test statistic:
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