Complete the following hypothesis test. Ho:o < 4.36 Ha:o > 4.36 (Claim) s = 3.6 , a = 0.10 and n = 29 Find a. The Critical Value x = b. The Test Statistic x = %3D The Chi-Square Test Statistic Degrees of Freedom (n – 1)s² n - 1 C. Is this a left, right or two tailed test? O right-tailed O left-tailed O two-tailed
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We have given that
Sample size n= 29 , s2 =3.6 , sigma^2 =4.36
Level of significance =alpha =0.10
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- Test the claim that the mean GPA of night students is larger than 2 at the 0.10 significance level. The null and alternative hypothesis would be: Họ :p = 0.5 Họ:µ = 2 Ho:u 0.5 Ho:p 2 H1 :p + 0.5 H1:µ # 2 H1:µ > 2 H1:p 0.5 H1:µ < 2 The test is: left-tailed right-tailed two-tailed Based on a sample of 25 people, the sample mean GPA was 2.02 with a standard deviation of 0.05 The test statistic is: (to 2 decimals) The p-value is: (to 2 decimals) Based on this we: Reject the null hypothesis Fail to reject the null hypothesisTest the claim that the proportion of people who own cats is smaller than 10% at the 0.025 significance level. The null and alternative hypothesis would be: Ho p ≤ 0.1 Ho:μ = 0.1 Ho:p 0.1 H₁:p> 0.1 H₁:μ 0.1 H₁:p ‡ 0.1 O The test is: = left-tailed two-tailed right-tailed Based on this we: Based on a sample of 600 people, 5% owned cats The p-value is: Ho≥ 0.1 Ho: ≤0.1 Ho:p> 0.1 H₁:μ 0.1 H₁:p < 0.1 O O Reject the null hypothesis O Fail to reject the null hypothesis (to 2 decimals)Z-Test Z-Test Use the calculator displays to the right to make a decision to reject or fail to reject the null hypothesis at a significance level of α = 0.05. Inpt: Data Statsu #70 Ho:70 o:3.75 x:68.75 In:40 H:Ho Ho Ho Calculate Draw Choose the correct answer below. O A. Since the P-value is greater than x, fail to reject the null hypothesis. OB. Since the P-value is greater than x, reject the null hypothesis. OC. Since the P-value is less than x, reject the null hypothesis. O D. Since the P-value is less than x, fail to reject the null hypothesis. z=2.10818511 p=0.03501498 x = 68.75 n=40
- q7This Question: 4 pts 1 of 9 (0 complete) This Quiz: 45 pts possible Assume a significance level of a = 0.05 and use the given information to complete parts (a) and (b) below. Original claim: More than 54% of adults would erase all of their personal information online if they could. The hypothesis test results in a P-value of 0.1661. a. State a conclusion about the null hypothesis. (Reject H, or fail to reject Ho.) Choose the correct answer below. A. Reject Ho because the P-value is greater than a. B. Reject Ho because the P-value is less than or equal to a. C. Fail to reject H, because the P-value is less than or equal to a. O D. Fail to reject H, because the P-value is greater than a. b. Without using technical terms, state a final conclusion that addresses the original claim. Which of the following is the correct conclusion? A. There is sufficient evidence to support the claim that the percentage of adults that would erase all of their personal information online if they could is…Test the claim that the mean GPA of night students is larger than 2.6 at the 0.01 significance level. The null and alternative hypothesis would be: Ho:p > 0.65 Ho:p 2.6 H1:p 0.65 H1:p+ 0.65 H1:µ > 2.6 H1:µ + 2.6 H:µ < 2.6 %3D The test is: two-tailed right-tailed left-tailed Based on a sample of 65 people, the sample mean GPA was 2.63 with a standard deviation of 0.07 The test statistic is: (to 2 decimals) The p-value is: (to 2 decimals) Based on this we: O Fail to reject the null hypothesis O Reject the null hypothesis
- Test the claim that the mean GPA of night students is larger than 2.4 at the 0.025 significance level. The null and alternative hypothesis would be: Ho: µ = — 2.4 Но:р 2.4 Н: + 2.4 H:p> 0.6 H:д> 2.4 Hi:p #0.6 Hi:p < 0.6 Н:р < 2.4 The test is: left-tailed two-tailed right-tailed Based on a sample of 65 people, the sample mean GPA was 2.41 with a standard deviation of 0.03 The test statistic is: (to 2 decimals) The p-value is: (to 2 decimals) Based on this we: O Reject the null hypothesis O Fail to reject the null hypothesisCan you help me with this part?3/ a. Test the claim that the contents of cans of diet soda have weights with a mean that is less than the mean for the regular soda. What are the null and alternative hypotheses? The test statistic, t, The P-value is State the conclusion for the test. A. Fail to reject the null hypothesis. There is sufficient evidence to support the claim that the cans of diet soda have mean weights that are lower than the mean weight for the regular soda. B. Reject the null hypothesis. There is not sufficient evidence to support the claim that the cans of diet soda have mean weights that are lower than the mean weight for the regular soda. C. Fail to reject the null hypothesis. There is not sufficient evidence to support the claim that the cans of diet soda have mean weights that are lower than the mean weight for the regular soda. D. Reject the null hypothesis. There is sufficient evidence to support the claim that the cans of diet soda have mean weights…
- The test statistic in a left-tailed test is z=−0.59. Determine the P-value and decide whether, at the 10% significance level, the data provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis. The P-value is ___________ (Round to four decimal places as needed.) This P-value ________________ (provides, does not provide) sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis because it is_____________(greater than, less than) the significance level.8. According to law enforcement records, the proportion of all drivers who have received at least one speeding ticket is 0.20. Let’s say we choose a large random sample of drivers and determine the proportion who have received at least one speeding ticket. We know, if the sampling method is repeated, that the sample proportion will vary from sample to sample. In fact, if we look at the sampling distribution in this case, we will see a distribution that is Normal in shape, with a mean (or center) of 0.20 and a standard deviation of 0.036. From this information, we know the middle 68% of this distribution will be between approximately what two values? 1. 0.19 and 0.21 2. 0.152 and 0.248 3. 0.1996 and 0.2004 4. 0.176 and 0.224 5. 0.164 and 0.236