1)Test of older baseballs showed that when dropped 22 ft. onto a concrete surface, they bounced an average of 235.2 cm. In a test of 50 new baseballs, the bounce height had a mean of 235.4 cm. Assume that the standard deviation of bounce heights of all new baseballs is 4.5 cm. Use a 0.01 significance level to test the claim that the new baseballs have bounce heights with a mean different from 235.2 cm. Are the new baseballs different? What are the null and alternative hypotheses? What is the value of the test statistic? Identify the critical value(s) of Z. Z= 2) Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Identify the null and alternative hypotheses, test statistic, P-value, critical value(s), and state the final conclusion that addresses the original claim.
Answer:
Null Hypothesis:
The bounce height of new baseballs when dropped at 22 ft. onto a concrete surface has a mean that is not significantly different from 235.2 cm or Ho: = 235.2 cm
Alternative Hypothesis:
The bounce height of new baseballs when dropped at 22 ft. onto a concrete surface has a mean that is significantly different from 235.2 cm or H1: ≠ 235.2 cm
Test at alpha = 0.01 level (2 tailed)
Using z test for one sample mean, (since the population standard deviation is known)
The two tailed critical z values are z = 2.5758
Test Statistic:
Z = (x – ) / (/n)
= (235.4 – 235.2) / ( 4.5 / 50)
= 0.3143
P value: = 0.7533
Decision:
Since the computed test statistic (0.3143) is not greater than z = 2.5758, then we failed to reject the null hypothesis:
Conclusion:
There is no sufficient statistical evidence to support the claim that the bounce height of new baseballs when dropped at 22 ft. onto a concrete surface has a mean that is significantly different from 235.2 cm or H1: ≠ 235.2 cm at alpha = 0.01 (2 tailed)
2)A simple random sample of pages from a dictionary is obtained. Listed below are the numbers of word defined on those pages. Given that this dictionary has 1459 pages with defined words, the claim that there are more than 70,000 defined words is the same as the claim that the mean number of defined words on a page is greater than 48.0. Use a 0.10 significance level to test this claim. What does the result suggest about the claim that there are more than 70,000 defined words in the dictionary? 80 48 57 67 65 75 44 42 94 98 what are the null and alternative hypotheses? Identify the test statistic. Identify the P-value Identify the critical value(s). State the final conclusion that addresses the original claim. Does the result suggest that there are more than 70,000 defined words in the dictionary?
Answer:
Null Hypothesis:
The mean number of words defined on the pages of a dictionary isnot significantly more than 70,000 or Ho: = 48
Alternative Hypothesis:
The mean number of words defined on the pages of a dictionary is significantly more than 70,000 or H1: 48
Test at alpha = 0.10level (1 tailed)
Using t test for one sample mean, (since the population standard deviation is not known)
The 1 tailed critical t value at df = n – 1 = 10 – 1 = 9 is t = 1.383
Test Statistic:
8048
57
67
65
75
44
42
94
98
mean / 67.00
sd / 19.84
t= (x – ) / (s/n)
= (67 – 48) / ( 19.84 / 10)
= 3.0284
P value: = 0.0071
Decision:
Since the computed test statistic (3.0284) is greater than t = 1.383, then we reject the null hypothesis:
Conclusion:
There is sufficient statistical evidence to support the claim that Themean number of words defined on the pages of a dictionary is significantly more than 70,000 or H1: 48 at alpha = 0.10 (1 tailed)
3) Use the given information to find the minimum sample size requires to estimate an unknown population mean μ. How many adults must be randomly selected to estimate the mean FICO (credit rating) score of working adults in a country? We want 90% confidence that the sample mean is within 4 points of the population mean, and the population standard deviation is 61. The minimum sample size is ______adults.
Answer:
n = (z * / e)2
= (1.6449 * 61 / 4)2
= (629.24 or rounded up to) 630
4) Claim: The mean IQ score of the statistics professors is less than 122. Sample Data: n=19, x=121, s=13. The significance level is 0.05 A) What is the test Statistic? = ____ Round to 3 decimal places B) What is the critical Values= ___ (use the t distribution table)Round to 3 decimal places as needed, Use a comma to separate answers as needed C)What is the P-Value = ___ Round to four decimal places as needed D) What is the final Conclusion:
Answer:
A)
Test statistic:
t= (x – ) / (s/n)
= (121 – 122) / ( 13/ 19)
= –0.335
B)
Critical t value: (df = 19 – 1 = 18)
t = –1.734
C)
P value = 0.3707
D)
Conclusion:
Do not reject the null hypothesis (since the p value is not less than 0.05)
5) In a test of effectiveness of garlic for lowering cholesterol, 47 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes in their levels of LDL cholesterol (in mg/dL) have a mean of 3.2 and a standard deviation of 15.3. Complete parts (a) and (b) below. A. What is the best point estimate of the population mean net change in LDL cholesterol after the garlic treatment? The best point estimate is ___ mg/dL. (Type an integer or a decimal.) B. Construct a 90% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol? What is the confidence interval estimate of the population mean u (mew)? __ mg/dL< u <____ mg/dL (Round to two decimal places as needed.) What does the confidence interval suggest about the effectiveness of the treatment?
Answer:
A)Best point estimate = 3.2 mg/dL
B)90% Confidence interval for the mean = sample mean Z * s / n
= 3.2 1.6449 * 15.3 / 47
= 3.2 3.67
= –0.47 < < 6.87
C.) Since 0 is included in the 90% confidence interval, then, this suggest that the garlic is effective in reducing LDL cholesterol.
D.) The confidence interval suggest the effectiveness of the treatment.
6) A simple random sample of 60 screws supplied by a certain manufacturer is obtained, and the length of each screw in measured. The sample mean is found to be 0.622 in. Assume that the standard deviation of all such length is 0.011 in, and use a 0.05 significance level to test the claim that the screws have a mean length equal to 5/8 in. (or 0.625 in), as indicated on the package labels. Do the screw lengths appear to be consistent with the package label? What are the null and alternative hypotheses? What us the value of the test statistic? The P-value is?
Answer:
Null Hypothesis:
The mean length of screws supplied by a certain manufacturer is not significantly different from 0.625 in or Ho: = 0.625 in.
Alternative Hypothesis:
The mean length of screws supplied by a certain manufactureris significantly different from 0.625 in or H1: ≠0.625 in.
Test at alpha = 0.05level (2 tailed)
Using z test for one sample mean, (since the population standard deviation is known)
The two tailed critical z values are z = 1.96
Test Statistic:
Z = (x – ) / (/n)
= (0.622 – 0.625) / ( 0.011 / 60)
= –2.1125
P value: = 0.0346
Decision:
Since the computed test statistic (–2.1125) is less than z = –1.96, then we reject the null hypothesis:
Conclusion:
There is sufficient statistical evidence to support the claim that the mean length of screws supplied by a certain manufactureris significantly different from 0.625 in or H1: ≠0.625 in at alpha = 0.05 (2 tailed)
The screw lengths appear to be not consistent with the package label
7) Trails in an experiment with a polygraph include 96 results that include 24 cases of wrong results and 72 correct results. Use a 0.05 significance level to test the claim that such polygraph results are correct less than 80% of the time. Identify the null hypothesis, test statistic, p-value, conclusion about the null hypothesis, and final conclusion that address the original claim. Use the P-method. Use the normal distribution as an approximation of the binomial distribution. What are the null and alternative hypotheses? The test statistic is z= The P- value is?
Answer:
Null Hypothesis:
The percent of the polygraph results which are correct is not significantly less than 80% or Ho: P = 80%
Alternative Hypothesis:
The percent of the polygraph results which are correct is significantly less than 80% or H1: P80%
Test at alpha = 0.05level (1 tailed)
Using z test for one sample proportion,
The 1 tailed critical z value is z = –1.6449
Test Statistic:
z= (p – P) / (PQ/n)
= (72/96 – 0.8) / (0.8*0.2/96)
= –1.2247
P value: = 0.1103
Decision:
Since the computed p value is (0.1103) is not less than the alpha value of 0.05, then we failed to reject the null hypothesis:
Conclusion:
There is no sufficient statistical evidence to support the claim that The percent of the polygraph results which are correct is significantly less than 80% or H1: P80% at alpha = 0.05 (1 tailed)