In 2010, Dr. Bob decided to gather research on the type of
disorders that present among his patients. His data collection
resulted in the following breakdown of patients by disorder: 54.9%
Schizophrenia; 21.1% Major Depression; 7.9% Obsessive-Compulsive
Disorder; 4.5% Anxiety Disorder; 2.9% Personality Disorder; 8.8%
Other. Information was collected from a random sample 0f 300
patients in 2018 to determine whether or not the data has changed
significantly. The sample data is given in the table below. At the
α=0.05 level of significance, test the claim that the disorder
breakdown of patients at Dr. Bob's hospital has not changed
significantly since 2010.
Which would be correct hypotheses for this test?
Type of disorder per patient in sample:
| Disorder | Count |
|---|---|
| Schizophrenia | 145 |
| Major Dispression | 73 |
| Obsessive-Compulsive Disorder | 30 |
| Anxiety Disorder | 12 |
| Personality Disorder | 14 |
| Other | 26 |
Test Statistic:
Give the P-value:
Which is the correct result:
Which would be the appropriate conclusion?
In: Math
Sample mean: x̄ = 48.74
Sample standard deviation: s = 32.5857
Size of your sample: n = 50
What is your Point Estimate? (round each answer to at least 4 decimals)
For a 95% confidence interval
In: Math
please provide me below
just i want
z tables , t tables , chi square tables
both right tailed ,left tailed .,two tailed
high quality of images only,other wise it wont help me, it leads thumbdown.
thankyou chegg
In: Math
Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 14 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is x = 3.15 grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with σ = 0.22 gram.
(a)
Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error? (Round your answers to two decimal places.)
lower limitupper limitmargin of error
(b)
What conditions are necessary for your calculations? (Select all that apply.)
σ is knownσ is unknownn is largeuniform distribution of weightsnormal distribution of weights
(c)
Interpret your results in the context of this problem.
There is an 80% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region.There is a 20% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region. The probability to the true average weight of Allen's hummingbirds is equal to the sample mean.The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20.The probability that this interval contains the true average weight of Allen's hummingbirds is 0.80.
(d)
Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.10 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)
hummingbirds
In: Math
The following table was derived from a study of HIV patients, and the data reflect the number of subjects classified by their primary HIV risk factor and gender. Test if there is a relationship between HIV risk factor and gender using a 5% level of significance:
|
Gender |
Total |
||
|
HIV Risk Factor |
Male |
Female |
|
|
IV drug user |
24 |
40 |
64 |
|
Homosexual |
32 |
18 |
50 |
|
Other |
15 |
25 |
40 |
|
71 |
83 |
154 |
|
What type of chi-square test will you use (goodness of fit or
test of independence)?
What are your hypotheses?
H0:
HA:
Fill in the following table to calculate your test statistic:
|
IV Drug Use: |
Homosexual: |
Other: |
Total |
||
|
Male |
O = |
24 |
32 |
15 |
71 |
|
E = |
|||||
|
(O – E) = |
|||||
|
(O – E)2 / E = |
|||||
|
Female |
O = |
40 |
18 |
25 |
83 |
|
E = |
|||||
|
(O – E) = |
|||||
|
(O – E)2 / E = |
df =___________________
Critical value: ______________________
Conclusion: We _____________________ (reject / fail to reject) the
Null Hypothesis
Interpretation:
In: Math
Different varieties of the tropical flower Heliconia are fertilized by different species of hummingbirds. Over time, the lengths of the flowers and the form of the hummingbirds' beaks have evolved to match each other. Here are data on the lengths in millimeters of three varieties of these flowers on the island of Dominica. data332.dat
Do a complete analysis that includes description of the data and a significance test to compare the mean lengths of the flowers for the three species. (Round your answers for x to four decimal places, s to three decimal places, and s_(x^^\_) to three decimal places. Round your test statistic to two decimal places. Round your P-value to three decimal places.)
flower type n x^^\_ s s_(x^^\_)
H. bihai
H. caribaea red H.
caribaea yellow
F =
P =
variety length bihai 49.62 bihai 46.47 bihai 48.14 bihai 47.49 bihai 47.82 bihai 47.48 bihai 48.03 bihai 46.99 bihai 46.57 bihai 51.08 bihai 45.65 bihai 49.78 bihai 49.14 bihai 47.77 bihai 46.91 bihai 47.77 red 37.12 red 41.89 red 38.85 red 39.33 red 40.72 red 40.37 red 41.27 red 41.67 red 39.5 red 41.93 red 41.09 red 42.74 red 40.78 red 39.27 red 40.42 red 41.52 red 40.6 red 38.67 red 37.94 red 41.86 red 37.44 red 42.01 red 40.62 yellow 35.07 yellow 36.23 yellow 34.55 yellow 36.56 yellow 35.33 yellow 36.96 yellow 36.23 yellow 33.76 yellow 34.25 yellow 35.41 yellow 35.25 yellow 37.49 yellow 35.04 yellow 37.48 yellow 37.15
In: Math
Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 14 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is x = 3.15 grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with σ = 0.20 gram. (a) Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error? (Round your answers to two decimal places.) lower limit upper limit margin of error (b) What conditions are necessary for your calculations? (Select all that apply.) normal distribution of weights uniform distribution of weights n is large σ is known σ is unknown (c) Interpret your results in the context of this problem. The probability to the true average weight of Allen's hummingbirds is equal to the sample mean. The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20. There is an 80% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region. There is a 20% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region. The probability that this interval contains the true average weight of Allen's hummingbirds is 0.80. (d) Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.15 for the mean weights of the hummingbirds. (Round up to the nearest whole number.) ___hummingbirds
In: Math
In: Math
part 1
An independent measures study was conducted to determine whether a
new medication called "Byeblue" was being tested to see if it
lowered the level of depression patients experience. There were two
samples, one that took ByeBlue every day and one that took a
placebo every day. Each group had n = 30. What is the df value for
the t-statistic in this study?
a. 60
b. 58
c. 28
d. There is not enough information
part 2.
Engineers must consider the breadths of male heads when designing
helmets. The company researchers have determined that the
population of potential clientele have head breadths that are
normally distributed with a mean of 6.2-in and a standard deviation
of 0.8-in. Due to financial constraints, the helmets will be
designed to fit all men except those with head breadths that are in
the smallest 3.9% or largest 3.9%.
What is the minimum head breadth that will fit the clientele?
min =
What is the maximum head breadth that will fit the clientele?
max =
Do not round your answer.
In: Math
In how many ways can a five horse race end, allowing for the possibility that
two horses tie?
In: Math
A phone manufacturer wants to compete in the touch screen phone market. Management understands that the leading product has a less than desirable battery life. They aim to compete with a new touch phone that is guaranteed to have a battery life more than two hours longer than the leading product. A recent sample of 65 units of the leading product provides a mean battery life of 5 hours and 39 minutes with a standard deviation of 92 minutes. A similar analysis of 51 units of the new product results in a mean battery life of 7 hours and 53 minutes and a standard deviation of 83 minutes. It is not reasonable to assume that the population variances of the two products are equal. Use Table 2. Sample 1 is from the population of new phones and Sample 2 is from the population of old phones. All times are converted into minutes. Let new products and leading products represent population 1 and population 2, respectively. a. Set up the hypotheses to test if the new product has a battery life more than two hours longer than the leading product. H0: μ1 − μ2 = 120; HA: μ1 − μ2 ≠ 120 H0: μ1 − μ2 ≥ 120; HA: μ1 − μ2 < 120 H0: μ1 − μ2 ≤ 120; HA: μ1 − μ2 > 120 b-1. Calculate the value of the test statistic. (Round all intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) Test statistic b-2. Implement the test at the 5% significance level using the critical value approach. Do not reject H0; there is no evidence that the battery life of the new product is more than two hours longer than the leading product. Reject H0; there is no evidence that the battery life of the new product is more than two hours longer than the leading product. Do not reject H0; there is evidence that the battery life of the new product is more than two hours longer than the leading product. Reject H0; there is evidence that the battery life of the new product is more than two hours longer than the leading product.
In: Math
In a survey, 38% of the respondents stated that they talk to their pets on the telephone. A veterinarian believed this result to be too high, so he randomly selected 150 pet owners and discovered that 53 of them spoke to their pet on the telephone. Does the vet have a right to be skeptical? use the confidence interval .1 level of significance. a) because np0(1-p0)= blank (=, not equal, greater, or less than) 10, the sample size is (blank- less or greater) 5% of the population size and the sample (blank-) the requirements for testing the hypothesis (blank- are or are not) satisfied. b)what are the null and alternative hypotheses? c)determine the test statistic. d)determine the critical values. e) does the vet have a right to be skeptical?
In: Math
In: Math
Many studies have suggested that there is a link between exercise and healthy bones. Exercise stresses the bones and this causes them to get stronger. One study examined the effect of jumping on the bone density of growing rats. There were three treatments: a control with no jumping, a low-jump condition (the jump height was 30 centimeters), and a high-jump condition (60 centimeters). After 8 weeks of 10 jumps per day, 5 days per week, the bone density of the rats (expressed in mg/cm3 ) was measured. Here are the data. data126.dat
(a) Make a table giving the sample size, mean, and standard deviation for each group of rats. Consider whether or not it is reasonable to pool the variances. (Round your answers for x, s, and s_(x^^\_) to one decimal place.)
Group n x^^\_ s s_(x^^\_)
Control
Low jump
High jump
(b) Run the analysis of variance. Report the F statistic with its degrees of freedom and P-value. What do you conclude? (Round your test statistic to two decimal places and your P-value to three decimal places.)
F =
P =
Conclusion: There is no? or a? statistically significant difference between the three treatment means at the α = .05 level.
obs group g density 1 Control 1 565 2 Control 1 598 3 Control 1 611 4 Control 1 601 5 Control 1 623 6 Control 1 607 7 Control 1 595 8 Control 1 649 9 Control 1 620 10 Control 1 576 11 Lowjump 2 629 12 Lowjump 2 645 13 Lowjump 2 626 14 Lowjump 2 653 15 Lowjump 2 633 16 Lowjump 2 639 17 Lowjump 2 624 18 Lowjump 2 639 19 Lowjump 2 643 20 Lowjump 2 622 21 Highjump 3 619 22 Highjump 3 614 23 Highjump 3 606 24 Highjump 3 608 25 Highjump 3 615 26 Highjump 3 608 27 Highjump 3 620 28 Highjump 3 619 29 Highjump 3 597 30 Highjump 3 593
In: Math
Lester Hollar is vice president for human resources for a large manufacturing company. In recent years, he has noticed an increase in absenteeism that he thinks is related to the general health of the employees. Years ago, in an attempt to improve the situation, he began a fitness program in which employees exercise during their lunch hour. To evaluate the program, he selected a random sample of eight participants and found the number of days each was absent in the six months before the exercise program began and in the last six months. Below are the results. Use a 0.05 significance level and determine if it is reasonable to conclude that the number of absences has decline? Use this information to solve the following questions.
A. What is the null hypothesis statement for this problem?
B. What is the alternative hypothesis statement for this problem?
C. What is alpha for this analysis?
D. What is the most appropriate test for this problem? (choose one of the following)
a. t-Test: Paired Two Sample for Means
b. t-Test: Two-Sampled Assuming Equal Variances
c. t-Test: Two-Sample Assuming Unequal Variances
d. z-Test: Two Sample for Means
E. What is the value of the test statistic for the most appropriate analysis?
F. What is the lower bound value of the critical statistic? If one does not exist (i.e. is not applicable for this type analysis), document N/A as your response.
G. What is the upper bound value of the critical statistic? If one does not exist (i.e. is not applicable for this type analysis), document N/A as your response.
H. Is it reasonable to conclude that the number of absences has decline? (choose one of the following)
a. Yes
b. No
I. What is the p-value for this analysis? (Hint: Use this value to double check your conclusion)
| Employee | Before | After |
| 1 | 6 | 5 |
| 2 | 6 | 2 |
| 3 | 7 | 1 |
| 4 | 7 | 3 |
| 5 | 4 | 3 |
| 6 | 3 | 6 |
| 7 | 5 | 3 |
| 8 | 6 | 7 |
Show all work with the right formulas
In: Math