Question

Chi-Square test for a normal distribution.

The South Bend Station of the South Shore Railroad is having financial difficulties. The daily number of riders is believed to be normally distributed and the Railroad needs to know the distribution for scheduling. A sample of the number of riders for a 30 day period resulted in a sample mean of 24.5 and the sample standard deviation was 3. The actual data for the 30 day period is given below. Does it fit a normal distribution? Use a 10% significance level. 18, 25, 26, 27, 25, 20, 22, 23, 25, 25, 28, 22, 27, 20, 19, 31, 26, 27, 25, 24, 21, 29, 28, 22, 24, 24, 25, 25, 26, 26.

a) Ho :

Ha :

b) Determine the intervals: (use the Empirical Rule)

c) Bin the data using the intervals: Interval Tally fi E (n∙pi)

d) Calculate the test statistic:

e) Critical Value from the table:

f) Conclusion:

Answer #1

Ho : given data follow normal distribution

Ha: given data does not follow normal distribution

Expected value are in Ei column

TS = 3.2692

critical value for 10% alpha = 7.7794

since TS < critical value

we fail to reject the null hypothesis

The following data are believed to have come from a normal
distribution. Use the goodness of fit test and = .05 to
test this claim. Use Table 12.4.
18
23
21
24
18
22
18
22
19
13
10
20
17
20
20
20
17
14
24
23
22
43
28
27
26
29
28
33
22
28
Using six classes, calculate the value of the test statistic (to
2 decimals).______
The p value is ? _________
Conclusion? Cannot reject assumption...

in performing a chi-square goodness of fit test for a
normal distribution, a researcher wants to make sure that all of
the expected cell frequencies are at least five. the sample is
divided into 7 intervals. thw seco d through the sixth intervals
all have expected cell frequencies of at least five. the first and
the last interval have expected cell frequencies of 1.5 each. after
adjusting the number of intervals, the freedom for the chi-square
statistic is

1.
The test statistic for goodness of fit has a chi-square
distribution with k - 1 degrees of freedom provided that
the expected frequencies for all categories are
a.
k or more.
b.
10 or more.
c.
5 or more.
d.
2k.
2. The owner of a car wash wants to see if the arrival rate of
cars follows a Poisson distribution. In order to test the
assumption of a Poisson distribution, a random sample of 150
ten-minute intervals was...

You will perform a Chi-Square test and an
ANOVA test. For each hypothesis test make sure to
report the following steps:
Identify the null hypothesis, Ho, and the alternative
hypothesis, Ha.
Determine whether the hypothesis test is left-tailed,
right-tailed, or two-tailed.
Find the critical value(s) and identify the rejection
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Find the appropriate standardized test statistic. If convenient,
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Interpret the decision in the context of the original...

Is there an association between patient BMI and cholesterol
level at screening? Consider overweight, normal and underweight
categories for BMI. Do a Chi-Square. I am having trouble finding
the test statistics. I picked the level of significance to be
5%.
Patient ID
Drug
Sex
Age
Height
BMI
Weight
Cholesterol
1
B
F
22
67.13
21.52787
138
197
2
B
M
22
63
20.36911
115
181
3
B
F
22
72
25.76582
190
190
4
A
M
22
69
16.98068
115...

The National Endowment for the Humanities sponsors summer
institutes to improve the skills of high school teachers of foreign
languages. At the beginning of the period, the teachers were given
the Modern Language Association’s listening test of understanding
of spoken French. A random sample of 15 teachers’ test score was
collected. After 4 weeks of immersion in French in and out of
class, the listening test was given again. From the second test
results, another 15 3 teachers’ test scores...

6. A researcher wishes to test the effect of early discharge of
cardiac patients on their ability to cope at home. The researcher
administers the coping scale to 10 cardiac patients who were
discharged early and 10 patients who stayed in the hospital for a
longer period of time. Higher scores indicate better coping. Based
on the data given in the table below, can you conclude that early
discharge has an effect on cardiac patients’ on ability to cope at...

A research council wants to estimate the mean length of time (in
minutes) that the average U.S. adult spends watching television
using digital video recorders (DVR’s) each day. To determine the
estimate, the research council takes random samples of 35 U.S.
adults and obtains the following times in minutes.
24
27
26
29
33
21
18
24
23
34
17
15
19
23
25
29
36
19
18
22
16
45
32
12
24
35
14
40
30
19
14...

A research council wants to estimate the mean length of time (in
minutes) that the average U.S. adult spends watching television
using digital video recorders (DVR’s) each day. To determine the
estimate, the research council takes random samples of 35 U.S.
adults and obtains the following times in minutes.
24
27
26
29
33
21
18
24
23
34
17
15
19
23
25
29
36
19
18
22
16
45
32
12
24
35
14
40
30
19
14...

A research council wants to estimate the mean length of time (in
minutes) that the average U.S. adult spends watching television
using digital video recorders (DVR’s) each day. To determine the
estimate, the research council takes random samples of 35 U.S.
adults and obtains the following times in minutes.
24
27
26
29
33
21
18
24
23
34
17
15
19
23
25
29
36
19
18
22
16
45
32
12
24
35
14
40
30
19
14...

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