3.
5. In this case, the mean (μ) of the sampling distribution will be the same as the mean of the population,
which is 75 points.
The standard deviation (σ) of the sampling distribution (also known as the standard error) can be
calculated using the formula:
σ
σ X=
√n
where:
σ is the standard deviation of the population (3 points in this case), and
n is the size of the sample (36 weeks in this case).
So, the standard deviation of the sampling distribution would be:
3
σ X= =0.5
√ 36
So, the sampling distribution of the mean in this situation is a normal distribution with a mean (μ) of 75
points and a standard deviation (σ) of 0.5 points. This distribution describes how the mean test score
varies from week to week.
6. In this case, the mean (μ) of the sampling distribution will be the same as the mean of the population,
which is 40 centimeters.
The standard deviation (σ) of the sampling distribution (also known as the standard error) can be
calculated using the formula:
σ
σ X=
√n
where:
σ is the standard deviation of the population (7 centimeters in this case), and
n is the size of the sample (90 days in this case).
So, the standard deviation of the sampling distribution would be:
7
σ X= ≈ 0.74
√ 90
So, the sampling distribution of the mean in this situation is a normal distribution with a mean (μ) of 40
centimeters and a standard deviation (σ) of approximately 0.74 centimeters. This distribution describes
how the mean diameter of the watermelons varies from day to day. However, it’s important to note that
this is an approximation because the original population is log-normally distributed, not normally
distributed. The Central Limit Theorem guarantees that the sampling distribution will be approximately
normal for large sample sizes, regardless of the shape of the population distribution. In this case, with a
sample size of 90, we can consider this approximation to be reasonably accurate.
7. To answer this question, we can perform a one-sample t-test. The null hypothesis (H0) is that the mean
score for the class is 70, and the alternative hypothesis (H1) is that the mean score for the class is above
70.
Ho: The mean score for the class is 70.
H1: The mean score of the class is >70.
First, let’s calculate the sample mean (x̄ ) and the sample standard deviation (s). The sample mean is the
sum of the scores divided by the number of scores, and the sample standard deviation is the square root
of the variance.
The scores are: 62, 92, 75, 68, 83, 95.
So, the sample mean (x̄ ) is:
62+92+75+68+ 83+95
x́= ≈79.17
6
The sample standard deviation (s) can be calculated as follows:
1. Subtract the mean from each score to get the deviation of each score.
2. Square each deviation to get the squared deviations.
3. Sum up all the squared deviations.
4. Divide the sum by the number of scores minus 1 (this is the variance).
5. Take the square root of the variance to get the standard deviation.
After calculating, we find that the sample standard deviation (s) is approximately 12.70.
Next, we calculate the t-score using the formula:
x́−μ
t=
s/√n
where:
x̄ is the sample mean,
μ is the population mean (70 in this case),
s is the sample standard deviation, and
n is the sample size (6 in this case).
So, the t-score is:
79.17−70
t= ≈ 1.69
12.70
√6
Finally, we compare the calculated t-score with the critical t-value from the t-distribution table. For a 90%
confidence level and 5 degrees of freedom (n-1), the critical t-value is approximately 1.476. Since our
calculated t-score (1.69) is greater than the critical t-value (1.476), we reject the null hypothesis.
Therefore, the professor can have 90% confidence that the mean score for the class on the test would be
above 70. However, it’s important to note that this is a statistical inference based on a small sample size,
and actual results may vary. The professor might want to increase the sample size for a more accurate
estimate.
8. To perform a paired t-test, we need to calculate the mean and standard deviation of the differences,
and then use these values to calculate the t-statistic. The formula for the t-statistic in a paired t-test is:
d́
t=
sd
√n
where:
dˉ is the mean of the differences,
sd is the standard deviation of the differences, and
n is the number of pairs.
The differences in weights are: 13, 10, -9, 19, 23, 13, 7, 1, 21, 15, 6, 11, 5, 14, 15.
So, the mean of the differences (dˉ) is:
13+10−9+19+23+ 13+7+1+21+15+6 +11+5+14 +15
d́= ≈ 10.6
15
The standard deviation of the differences (sd) can be calculated as follows:
1. Subtract the mean from each difference to get the deviation of each difference.
2. Square each deviation to get the squared deviations.
3. Sum up all the squared deviations.
4. Divide the sum by the number of differences minus 1 (this is the variance).
5. Take the square root of the variance to get the standard deviation.
After calculating, we find that the standard deviation of the differences (sd) is approximately 8.8.
So, the t-statistic is:
10.6
t= ≈ 4.5
8.8
√ 15
Finally, we compare the calculated t-statistic with the critical t-value from the t-distribution table. For a
two-tailed test with a 0.05 significance level and 14 degrees of freedom (n-1), the critical t-value is
approximately 2.145. Since our calculated t-statistic (4.5) is greater than the critical t-value (2.145), we
reject the null hypothesis.
Therefore, we can conclude that there is a statistically significant difference in the weights of the
individuals before and after participating in the weight loss program. This suggests that the program is
effective in helping clients lose weight. However, it’s important to note that this is a statistical inference
based on a sample, and actual results may vary. The farmer might want to increase the sample size for a
more accurate estimate. Also, other factors such as diet, exercise, and individual health conditions could
also affect the results.
9. To determine the sample size needed to estimate the mean height within one inch with 93%
confidence, we can use the formula for the confidence interval for a population mean:
( )
2
Z ⋅σ
n=
E
where:
n is the sample size,
Z is the Z-score corresponding to the desired confidence level,
σ is the standard deviation of the population, and
E is the desired margin of error.
In this case, the standard deviation (σ) is 2.5 inches, and the desired margin of error (E) is 1 inch. The Z-
score for a 93% confidence level is approximately 1.81 (you can find this value in a standard Z-table or
use a statistical calculator).
Substituting these values into the formula, we get:
( )
2
1.81 ⋅2.5
n= ≈ 20.3
1
Since we can’t measure a fraction of a person, we round up to the nearest whole number. Therefore, you
must measure at least 21 male students to estimate the mean height to within one inch with 93%
confidence. Please note that this is a statistical estimate, and actual results may vary. Also, this
calculation assumes that the heights of the students are normally distributed and that the sample is
randomly selected. If these assumptions are not met, a larger sample size may be needed.
10. To determine if the medication has an effect on intelligence, we can perform a one-sample t-test. The
null hypothesis (H0) is that the mean IQ of the population is 100, and the alternative hypothesis (H1) is
that the mean IQ of the population is not 100.
First, let’s calculate the sample mean (x̄ ) and the sample standard deviation (s). The sample mean is
given as 140.
The standard deviation (σ) of the population is given as 15. However, we don’t have the standard
deviation for the sample, so we’ll use the population standard deviation as an estimate. This is a common
practice when the sample size is large (n > 30), but since our sample size is exactly 30, it’s important to
note that this is an approximation.
Next, we calculate the t-score using the formula:
x́−μ
t=
σ
√n
where:
x̄ is the sample mean (140 in this case),
μ is the population mean (100 in this case),
σ is the standard deviation of the population (15 in this case), and
n is the size of the sample (30 in this case).
So, the t-score is:
140−100
t= ≈ 15.4
15
√30
Finally, we compare the calculated t-score with the critical t-value from the t-distribution table. For a two-
tailed test with a 0.05 significance level and 29 degrees of freedom (n-1), the critical t-value is
approximately 2.045. Since our calculated t-score (15.4) is much greater than the critical t-value (2.045),
we reject the null hypothesis.
Therefore, we can conclude that there is a statistically significant difference in the IQ scores of the
individuals before and after taking the medication. This suggests that the medication does have an effect
on intelligence. However, it’s important to note that this is a statistical inference based on a sample, and
actual results may vary. Also, other factors such as diet, exercise, and individual health conditions could
also affect the results. It’s also worth noting that an increase in IQ score doesn’t necessarily equate to an
increase in “intelligence,” as intelligence is a multi-faceted attribute that can’t be fully measured by an IQ
test. Finally, this statistical test assumes that the IQ scores are normally distributed; if this assumption is
not met, the results of the test may not be valid.
