a. null and alternate hypothesis? c.the p value of the test statistic? b. the va
ID: 3074008 • Letter: A
Question
a. null and alternate hypothesis? c.the p value of the test statistic?
b. the value of the test statistics? d.your conclusion about the null?
1. Past experience has shown that If the quality of teaching is similar in a school, the scores on a standardized test will have a variance of 400. The superintendent wants to know if there is a disparity in teaching quality and decides to investigate whether the variance of test scores has decreased. She samples 25 random students and finds a mean score of 85 with a variance of 200 Use =0.05. 2. A hospital director is told that 80% of the emergency room visitors are insured. The director wants to prove that the percentage of insured patients is under the expected percentage. A sample of 400 patients found that 300 were insured. Use a 99% level of confidenceExplanation / Answer
#1.
Below are the null and alternate hypothesis
H0: sigma^2 = 400
Ha: sigma^2 < 400
test stastics, chi-square = (25-1)*200/400 = 12
degree of feedom = 24
critical value of chi-square for alpha = 0.05 and df = 24 is 13.848
p-value = 0.0201
This is left tailed test and test statistics lie in the rejection region.
Hence we reject the null hypothesis.
There is significant evidence to conclude that the variance is less than 400.
#2.
Below are the null and alternate hypothesis
H0: p = 0.8
Ha: p < 0.8
pcap = 300/400 = 0.75
SE = sqrt(0.8*0.2/400) = 0.02
Test statistics, z = (0.75 - 0.8)/0.02 = -2.5
p-value = 0.0062
Here significance level is 0.01
We reject the null hypothesis.
There is significant evidence to conclude that teh percentage of insured patients is under 80%
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