Question

In: Statistics and Probability

Q1 A random sample of 36 values is drawn from a mound-shaped and symmetric distribution.

 

Q1)

A random sample of 36 values is drawn from a mound-shaped and symmetric distribution. The sample mean is 9 and the sample standard deviation is 2. Use a level of significance of 0.05 to conduct a two-tailed test of the claim that the population mean is 8.5.

(A) Is it appropriate to use a Student's t distribution? Explain.

a) Yes, because the x distribution is mound-shaped and symmetric and σ is unknown.

b) No, the x distribution is skewed left.

c) No, the x distribution is skewed right.

d) No, the x distribution is not symmetric.No, σ is known.


How many degrees of freedom do we use?


(B) What are the hypotheses?

a) H0: μ < 8.5; H1: μ = 8.5

b) H0: μ = 8.5; H1: μ ≠ 8.5

c) H0: μ > 8.5; H1: μ = 8.5

d) H0: μ = 8.5; H1: μ > 8.5

e) H0: μ = 8.5; H1: μ < 8.5


(C) Compute the t value of the sample test statistic. (Round your answer to three decimal places.)

t =



(D) Estimate the P-value for the test.

a) P-value > 0.250

b) 0.100 < P-value < 0.250

c) 0.050 < P-value < 0.100

d) 0.010 < P-value < 0.050

e) P-value < 0.010


(E) Do we reject or fail to reject H0?

a) At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.

b) At the α = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.

c) At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.

d) At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.


(F) Interpret the results.

a) There is sufficient evidence at the 0.05 level to reject the null hypothesis.

b) There is insufficient evidence at the 0.05 level to reject the null hypothesis.

Q2

Let x be a random variable that represents the pH of arterial plasma (i.e., acidity of the blood). For healthy adults, the mean of the x distribution is μ = 7.4.† A new drug for arthritis has been developed. However, it is thought that this drug may change blood pH. A random sample of 36 patients with arthritis took the drug for 3 months. Blood tests showed that x = 8.4 with sample standard deviation s = 2.7. Use a 5% level of significance to test the claim that the drug has changed (either way) the mean pH level of the blood.

(A) What is the level of significance?


State the null and alternate hypotheses.

a) H0: μ ≠ 7.4; H1: μ = 7.4

b) H0: μ = 7.4; H1: μ > 7.4    

c) H0: μ = 7.4; H1: μ < 7.4

d) H0: μ = 7.4; H1: μ ≠ 7.4

e) H0: μ > 7.4; H1: μ = 7.4


(B) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution.

a) The standard normal, since the sample size is large and σ is unknown.

b) The Student's t, since the sample size is large and σ is unknown.    

c) The standard normal, since the sample size is large and σ is known.

d) The Student's t, since the sample size is large and σ is known.


What is the value of the sample test statistic? (Round your answer to three decimal places.)


(C) Estimate the P-value.

a) P-value > 0.250

b) 0.100 < P-value < 0.250    

c) 0.050 < P-value < 0.100

d) 0.010 < P-value < 0.050

e) P-value < 0.010


Sketch the sampling distribution and show the area corresponding to the P-value.

Top left, bottom left, top right or bottom right?

   
   


(D) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level α?

a) At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.

b) At the α = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.    

c) At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.

d) At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.


(E) Interpret your conclusion in the context of the application.

a) There is sufficient evidence at the 0.05 level to conclude that the drug has changed the mean pH level of the blood.

b) There is insufficient evidence at the 0.05 level to conclude that the drug has changed the mean pH level of the blood.

Solutions

Expert Solution

1)

(A) Is it appropriate to use a Student's t distribution? Explain.

a) Yes, because the x distribution is mound-shaped and symmetric and σ is unknown.


How many degrees of freedom do we use?

df = n-1 = 36-1 = 35

(B) What are the hypotheses?

b) H0: μ = 8.5; H1: μ ≠ 8.5

(D) Estimate the P-value for the test.

a) P-value > 0.250

Here P-value is large.so we fail to reject the null hypothesis.


(E) Do we reject or fail to reject H0?

d) At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.


(F) Interpret the results.

b) There is insufficient evidence at the 0.05 level to reject the null hypothesis.

2)

(A) What is the level of significance?
0.05


State the null and alternate hypotheses.

d) H0: μ = 7.4; H1: μ ≠ 7.4


(B) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution.

b) The Student's t, since the sample size is large and σ is unknown.    

d) 0.010 < P-value < 0.050

P-value = 0.0328 < 0.05

So we reject Ho.

(D) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Is the data statistically significant at level α?

a) At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.


(E) Interpret your conclusion in the context of the application.

a) There is sufficient evidence at the 0.05 level to conclude that the drug has changed the mean pH level of the blood.


Related Solutions

A random sample of 16 values is drawn from a mound-shaped and symmetric distribution. The sample...
A random sample of 16 values is drawn from a mound-shaped and symmetric distribution. The sample mean is 12 and the sample standard deviation is 2. Use a level of significance of 0.05 to conduct a two-tailed test of the claim that the population mean is 11.5. (a) Is it appropriate to use a Student's t distribution? Explain. (Choose one of 5 listed below) Yes, because the x distribution is mound-shaped and symmetric and σ is unknown. No, the x...
A random sample of 16 values is drawn from a mound-shaped and symmetric distribution. The sample...
A random sample of 16 values is drawn from a mound-shaped and symmetric distribution. The sample mean is 8 and the sample standard deviation is 2. Use a level of significance of 0.05 to conduct a two-tailed test of the claim that the population mean is 7.5. (a) Is it appropriate to use a Student's t distribution? Explain. Yes, because the x distribution is mound-shaped and symmetric and σ is unknown.No, the x distribution is skewed left.      No, the x...
A random sample of 25 values is drawn from a mound-shaped and symmetric distribution. The sample...
A random sample of 25 values is drawn from a mound-shaped and symmetric distribution. The sample mean is 13 and the sample standard deviation is 2. Use a level of significance of 0.05 to conduct a two-tailed test of the claim that the population mean is 12.5. (a) Is it appropriate to use a Student's t distribution? Explain. Yes, because the x distribution is mound-shaped and symmetric and σ is unknown. No, the x distribution is skewed left. No, the...
A random sample of 25 values is drawn from a mound-shaped and symmetric distribution. The sample...
A random sample of 25 values is drawn from a mound-shaped and symmetric distribution. The sample mean is 9 and the sample standard deviation is 2. Use a level of significance of 0.05 to conduct a two-tailed test of the claim that the population mean is 8.5. (a) Is it appropriate to use a Student's t distribution? Explain. Yes, because the x distribution is mound-shaped and symmetric and σ is unknown. No, the x distribution is skewed left. No, the...
The distribution of the heights of the first grade students is mound – shaped and symmetric...
The distribution of the heights of the first grade students is mound – shaped and symmetric with the mean height of 140 cm and standard deviation of 5 cm. According Empirical rule What % of heights is between 125 and 155 cm What % of heights is between 135 and 150 cm What % of heights is less than 130 cm What % of heights is less than 130 cm
Suppose that the distribution of scores on an exam is mound shaped and approximately symmetric. The...
Suppose that the distribution of scores on an exam is mound shaped and approximately symmetric. The exam scores have a mean of 110 and the 16th percentile is 85. (Use the Empirical Rule.) (a) What is the 84th percentile? (b) What is the approximate value of the standard deviation of exam scores? (c) What is the z-score for an exam score of 90? (d) What percentile corresponds to an exam score of 160? % (e) Do you think there were...
Two populations have mound-shaped, symmetric distributions. A random sample of 16 measurements from the first population...
Two populations have mound-shaped, symmetric distributions. A random sample of 16 measurements from the first population had a sample mean of 20, with sample standard deviation 2. An independent random sample of 9 measurements from the second population had a sample mean of 19, with sample standard deviation 3. Test the claim that the population mean of the first population exceeds that of the second. Use a 5% level of significance. a) What distribution does the sample test statistic follow?...
1. Suppose that the distribution of scores on an exam is mound shaped and approximately symmetric....
1. Suppose that the distribution of scores on an exam is mound shaped and approximately symmetric. The exam scores have a mean of 110 and the 16th percentile is 85. a. What is the 84th percentile? b. What is the approximate value of the standard deviation of exam scores? c. What is the z-score for an exam score of 90? d. What percentile corresponds to an exam score of 160? e. Do you think there were many scores below 35?...
The distribution of ages of Oscar winning actors is roughly symmetric and mound shaped (in other...
The distribution of ages of Oscar winning actors is roughly symmetric and mound shaped (in other words, the ages for Oscar winning actors are normally distributed). The mean of this distribution is 44 years and the standard deviation of this distribution is 9 years. The 68-95-99.7 Rule Finding Percentiles a.What percent of actors received the Oscar after the age of 70? Draw the picture of the area that represents this probability. (You will notice that it is a right-tailed area.)...
The distribution of ages of Oscar winning actresses is roughly symmetric and mound shaped (in other...
The distribution of ages of Oscar winning actresses is roughly symmetric and mound shaped (in other words, the ages for Oscar winning actresses are normally distributed). The mean of this distribution is 36 years and the standard deviation of this distribution is 11 years.In 2013, Jennifer Lawrence and Daniel Day-Lewis won the Oscar for best leading actress and best leading actor respectively. Jennifer Lawrence was 22 years old and Daniel Day-Lewis was 55 years old a.Which is more unusual –...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT