null and alternative hypothesis
null and alternative hypothesis
NAME__________________ Score ______ / 50
STAT 200: Introduction to Statistics
Homework #5: Lesson 8, Sections 1-5 through Lesson 9, Sections 1-2
1. (2 points) A claim is made that when parents use the XSORT method of gender selection during in-vitro fertilization, the proportion of baby girls is greater than 0.5. The latest results show that among 945 babies born to couples using the XSORT method of gender selection, 879 were girls.
a. (1 point) Express the original claim in symbolic form.
b. (1 point) Identify the null and alternative hypothesis.
2. (10 points) A 0.05 significance level is used for a hypothesis test of the claim that when parents use the XSORT method of gender selection, the proportion of baby girls is different from 0.5. Assume that the data consists of 55 girls born in 100 births, so the sample statistic of 0.55 results in a z-score that is 1.00 standard deviation above 0.
a. (1 point) Identify the null and alternative hypothesis.
b. (1 point) What is the value of α?
c. (1 point) What is the sampling distribution of the sample statistic?
d. (1 point) Is this a two-tailed, left-tailed, or right-tailed test? Why?
e. (1 point) What is the value of the test statistic?
f. (2 points) What is the P-value?
g. (1 point) What is the critical value?
h. (1 point) What is the area of the critical region?
i. (1 point) What is the result of the hypothesis test (i.e. “Reject the Null Hypothesis” or “Fail to Reject the Null Hypothesis)? Why?
3. (8 points) The data set below contains data from a simple random sample of 100 M&Ms, 8 of which are brown (i.e. 8% or the proportion of 8 out of 100 are brown). Use a 0.05 significance level to test the claim of the Mars Candy Company that the percentage of brown M&Ms is equal to 13%.
Count
Red
Orange
Yellow
Brown
Blue
Green
1
0.751
0.735
0.883
0.696
0.881
0.925
2
0.841
0.895
0.769
0.876
0.863
0.914
3
0.856
0.865
0.859
0.855
0.775
0.881
4
0.799
0.864
0.784
0.806
0.854
0.865
5
0.966
0.852
0.824
0.840
0.810
0.865
6
0.859
0.866
0.858
0.868
0.858
1.015
7
0.857
0.859
0.848
0.859
0.818
0.876
8
0.942
0.838
0.851
0.982
0.868
0.809
9
0.873
0.863
0.803
0.865
10
0.809
0.888
0.932
0.848
11
0.890
0.925
0.842
0.940
12
0.878
0.793
0.832
0.833
13
0.905
0.977
0.807
0.845
14
0.850
0.841
0.852
15
0.830
0.932
0.778
16
0.856
0.833
0.814
17
0.842
0.881
0.791
18
0.778
0.818
0.810
19
0.786
0.864
0.881
20
0.853
0.825
21
0.864
0.855
22
0.873
0.942
23
0.880
0.825
24
0.882
0.869
25
0.931
0.912
26
0.887
27
0.886
a. (1 point) Identify the null and alternative hypothesis.
b. (1 point) Is this a two-tailed, left-tailed, or right-tailed test? Why?
c. (1 point) What is the value of the test statistic?
d. (2 points) What is the P-value?
e. (1 point) What is the critical value?
f. (1 point) What is the area of the critical region?
g. (1 point) What is the result of the hypothesis test (i.e. “Reject the Null Hypothesis” or “Fail to Reject the Null Hypothesis)? Why did you respond with this answer, and what does it mean?
4. (10 points) The data set in problem 3 presents a sample of 100 plain M&M candies that randomly selected (without replacement) from a bag which contained a total of 465 M&M candies. The weight of each M&M (in grams) is recorded in the table above and in the available Excel Data Set file. From this simple random sample, there were 19 green M&Ms with a mean of 0.8635 g and a standard deviation of 0.0570. Use a 0.05 significance level to test the claim that the mean weight of all M&Ms is equal to 0.8535, which is the mean weight required so that M&Ms have the weight printed on the packaged label.
a. (1 point) Identify the null and alternative hypothesis.
b. (1 point) Is this a two-tailed, left-tailed, or right-tailed test? Why?
c. (2 points) What is the value of the test statistic?
d. (2 points) What is the P-value?
e. (1 point) What is the critical value?
f. (1 point) What is the area of the critical region?
g. (1 point) What is the result of the hypothesis test (i.e. “Reject the Null Hypothesis” or “Fail to Reject the Null Hypothesis)? Why did you respond with this answer, and what does it mean?
h. (1 point) Do green M&Ms appear to have weights consistent with the package label?
5. (10 points) The Wechsler IQ test is designed so that the mean is 100 and the standard deviation is 15 for the population of normal adults. Listed below (and in the Excel Data File) are the IQ scores of randomly selected professional pilots. It is claimed that because the professional pilots are a more homogenous group than the general population (i.e. they are “more alike” than a random group of people), they have IQ scores with a standard deviation less than 15. Test the claim using a 0.05 significance level.
121
116
115
121
116
107
127
98
116
101
130
114
a. (1 point) Identify the null and alternative hypothesis.
b. (1 point) Is this a two-tailed, left-tailed, or right-tailed test? Why?
c. (2 points) What is the value of the test statistic?
d. (2 points) What is the critical value?
e. (2 points) What is the P-value?
f. (1 point) What is the area of the critical region?
g. (1 point) What is the result of the hypothesis test (i.e. “Reject the Null Hypothesis” or “Fail to Reject the Null Hypothesis)? Why did you respond with this answer, and what does it mean?
6. (10 points) Listed below (and in the available Excel Data Set file) are the PSAT and SAT scores from prospective college applicants. The scores were reported by subjects who responded to a request posted by the web site talk.collegconfidential.com. There is a claim that higher PSAT scores correlate to higher SAT scores. Use this data set to argue for or against that claim.
PSAT
183
207
167
206
197
142
193
176
SAT
2200
2040
1890
2380
2290
2070
2370
1980
a. (1 point) Identify the null and alternative hypothesis.
b. (1 point) Is this a two-tailed, left-tailed, or right-tailed test? Why?
c. (2 points) What is the value of the test statistic?
d. (2 points) What is the critical value?
e. (2 points) What is the P-value?
f. (1 point) What is the result of the hypothesis test (i.e. “Reject the Null Hypothesis” or “Fail to Reject the Null Hypothesis)? Why did you respond with this answer, and what does it mean?
g. (1 point) Is there anything about the data that might make the results questionable?
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