3)
Test statistic = (n - 1) s2 /
= (10 -1) * 28.4 / 27
= 9.47
4)
P-value = P(
21
> 34.7) = 0.0304
0.025 < p < 0.05
QUESTION 3 9.3: Inferences about variance and standard deviation. Calculate the value of the test statistic,...
QUESTION 4 Given the test statistic and degrees of freedom, find the p-value range (area to the right) that is the best choice. x2 = 34.7, df = 21 0.75 < P<0.90 0.05 <P<0.10 0.90 < P<0.95 0.025 <P<0.05 QUESTION 5 Given the test statistic and degrees of freedom, find the p-value range (area to the right) that is the best choice. x2 = 35.2, df = 24 0.90 < P<0.95 0.025 <P<0.05 0.75 <P<0.90 0.05 < P<0.10
Question 5 and 6
2 = 34.7, df = 21 0.75 <P<0.90 0.025 <P<0.05 0.90 < P<0.95 О 0.05<P<0.10 UESTIONS Given the test statistic and degrees of freedom, find the p-value range (area to the right) that is the best choice. X2=2.76, df = 6 O 0.025 <P<0.05 0.05<P<0.10 0.90 <P<0.95 0.75 <P<0.90 QUESTION 6 When making inferences concerning the mean difference using two dependent samples, it is necessary to calculate the standard deviation of the sample differences Calculate the...
2 = 34.7, df = 21 0.75 <P<0.90 0.025 <P<0.05 0.90 < P<0.95 О 0.05<P<0.10 UESTIONS Given the test statistic and degrees of freedom, find the p-value range (area to the right) that is the best choice. X2=2.76, df = 6 O 0.025 <P<0.05 0.05<P<0.10 0.90 <P<0.95 0.75 <P<0.90 QUESTION 6 When making inferences concerning the mean difference using two dependent samples, it is necessary to calculate the standard deviation of the sample differences Calculate the standard deviation of the...
59. A chi-square test of independence with 10 degrees of freedom results in a test statistic of 19.25. Using the chi-square table, which of the following is the most accurate statement that can be made about the p-value for this test? A: p-value < 0.025 B: 0.05 < p-value < 0.10 C: 0.10 < p-value < 0.20 D: 0.025 < p-value < 0.05
For the below output, what was the test used, the p-value and
the conclusion?
a.
An F test, with 1 and 94 degrees of freedom; p
= 0.000;reject the null
b.
A chi-square test with 4 degrees of freedom; p = 0.383; do not
reject the null.
c.
A t-test with 91 degrees of freedom; p =
0.000; reject the null
d.
An F test with 4 and 91 degrees of freedom; p
= 0.000; reject the null
lSummary of...
Pick 3 correct inferences about the following hypothesis test to determine whether the mean thickness of aluminum sheets equals 0.05. Assume a 5% level of significance was used. A. The null hypothesis is mean 0.05". B. This is a one-tailed test. C. The chance of a Type 1 error is 1.4%. D, 68% of sheets should be +-0.01" of Hypothesis Test (One-Sample) Sample Size Sample Mean Sample Std Dev Hypothesized Mean Data Set #1 100 0.04739 0.01043 0.05 0.05 0.00104...
3) For the hypothesis test H0: μ = 5 against H1: μ < 5 with variance unknown and n = 12, approximate the P-value for each of the following test statistics. a) t0 = 2.05 b) t0 = −1.84 c) t0 = 0.4 EXPECTED ANSWERS: a) 0.95 ≤ p ≤ 0.975 b) 0.025 ≤ p ≤ 0.05 c) 0.6 ≤ p ≤ 0.75
For the below output, what was the test used, the p-value and
the conclusion?
a.
An F test, with 1 and 94 degrees of freedom; p
= 0.000;reject the null
b.
A chi-square test with 4 degrees of freedom; p = 0.383; do not
reject the null.
c.
A t-test with 91 degrees of freedom; p =
0.000; reject the null
d.
An F test with 4 and 91 degrees of freedom; p
= 0.000; reject the null
lSummary of...
Consider the following summary statistics, calculated from two independent random samples taken from normally distributed populations. Sample 1 Sample 2 x¯1=20.92 x¯2=26.80 s21=2.89 s22=3.81 n1=19 n2=15 Test the null hypothesis H0:μ1=μ2against the alternative hypothesis HA:μ1<μ2. a) Calculate the test statistic for the Welch Approximate t procedure. Round your response to at least 3 decimal places. b) The Welch-Satterthwaite approximation to the degrees of freedom is given by df = 27.983055. Using this information, determine the range in which the p-value...
Consider the following summary statistics, calculated from two independent random samples taken from normally distributed populations. Sample 1 Sample 2 x¯1=20.08 x¯2=24.51 s21=2.05 s22=3.20 n1=19 n2=16 Test the null hypothesis H0:μ1=μ2against the alternative hypothesis HA:μ1<μ2. a) Calculate the test statistic for the Welch Approximate t procedure. Round your response to at least 3 decimal places. b) The Welch-Satterthwaite approximation to the degrees of freedom is given by df = 28.610808. Using this information, determine the range in which the p-value...