Interpreting computer output: The following output from MINITAB presents the results of a hypothesis test. Test of μ = 232 vs < 232 The assumed standard deviation = 42 N 51 Mean SE Mean 5.88 219.56 95% Upper Bound 229.23 Z P -2.12 0.017
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- Find the 98% confidence interval for the difference between two means based on this information about two samples. Assume independent samples from normal populations. (Use conservative degrees of freedom.) (Give your answers correct to two decimal places. Sample Number Mean Std. Dev. 1 27 36 30 2 30 25 22 Lower Limit Upper LimitA researcher is testing a hypothesis of a single mean. The critical t value for a = .05 and a one-tailed test is 2.0639. The observed t value from sample data is 1.742. The decision made by the researcher based on this information is to the null hypothesis. reject not reject redefine change the alternate hypothesis into restateFind the 98% confidence interval for the difference between two means based on this information about two samples. Assume independent samples from normal populations. (Use conservative degrees of freedom.) (Give your answers correct to two decimal places.) Sample Number Mean Std. Dev. 1 16 38 29 2 29 28 29 Lower Limit Upper Limit
- A sample mean is 137, sample size is 20, and population standard deviation is 14.2 are given. Ho: μ-132, H1: µ#132 and a significance level is 0.05. Use the P-value approach to test the hypothesis. O a. a. z = 2.57; P-value = 0.0101; reject null hypothesis b. z = 0.35; P-value = 0.7263; do not reject null hypothesis O c. z = 1.57; P-value = 0.0582; do not reject null hypothesis O d. z = 1.57; P-value = 0.1164; do not reject null hypothesisUnder what circumstances is a t statistic used instead of a z-score for a hypothesis test? Justin wants to know whether a commonly prescribed drug does improve the attention span of students with attention deficit disorder (ADD). He knows that the mean attention span for students with ADD who are not taking the drug is 2.3 minutes long. His sample of 12 students taking the drug yielded a mean of 4.6 minutes. Justin can find no information regarding σx , so he calculated s2x =1.96. Determine the critical region using a one-tailed test with alpha = .05. Conduct the hypothesis test (Do the math and compare the t-critical and t-obtained values). State your conclusions in terms of H0 (Should you reject the H0 or fail to reject/accept the H0). Based on your analysis, is there a relationship between the drug and attention span?J1
- II Sh) with Zcrit of Therefore, (a = .01). %66 Level of Confidence Fail to Reject Reject Reject 0.5% 0.5% Zcrit = -2.576 bubiy Z crit = +2.576 3.33 %3D Let's Practice Two-Tailed Z Test Blood glucose levels for obese patients have a mean of 100 with a standard deviation of 15. A researcher thinks that a diet high in raw corn starch will have a negative or positive effect on blood glucose levels. A sample of 30 patients who have tried the raw corn starch diet have a mean glucose level of 140. Test the hypothesis that the raw cornstarch had an effect at a significant level of 0.05 Step 1: Determine the null hypothesis %3D = 11 : °H +ri : "H Step 2: Two-tailed test Step 3: Select significance level a = Step 4: Do we do a Z test of T test? Step 5: Find the critical value Zerii from the t Table. Level of Confidence Draw and label the graph. Fail to Reject From the t Table. Zerit for a= 0.05, + 1.96 Reject Reject Note that a= 0.05 (5%) means 2.5% in each tail. This alpha level indicates the…The mean voltage and standard deviation of 10 batteries from each manufacturer were measured. The results are summarized in the following table. Manufacturer Sample mean voltage (millivolts) Sample standard deviation (millivolts) A 136 3 B 132 5 What type of hypothesis test should be performed? What is the test statistic? What is the number of degrees of freedom? Does sufficient evidence exist to support the claim that the voltage of the batteries made by the two manufacturers is different at the α=0.1 significance level?(c) Identify the standardized test statistic. Use technology. z=kjkj (Round to two decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis, and (e) interpret the decision in the context of the original claim. A. RejectH0. There is sufficient evidence to reject the claim that mean bulb life is at least 743 hours. B. Fail to reject H0. There is sufficient evidence to reject the claim that mean bulb life is at least 743 hours. C. Reject H0. There is not sufficient evidence to reject the claim that mean bulb life is at least 743 hours. D. Fail to reject H0. There is not sufficient evidence to reject the claim that mean bulb life is at least 743 hours.
- What is the name of this hypothesis test? Do male and female servers at Applebee's work the same number of hours? A random sample of 45 female servers worked an average of 36 hours per week, with a standard deviation of 2. A random sample of 62 male servers worked an average of 29.5 hours per week, with a standard deviation of 4. O Chi-Square Test for Independence O One Sample z Test OTwo Proportion z Test Z. OLinear Regression t Test O Chi-Square Goodness of Fit Test O Paired Samples t Test OTwo Independent Samples t Test O One Sample t Test O One Proportion z Test OANOVABIG Corporation advertises that its light bulbs have a mean lifetime, u, of 3000 hours. Suppose that we have reason to doubt this claim and decide to do a statistical test of the claim. We choose a random sample of light bulbs manufactured by BIG and find that the mean lifetime for this sample is 3160 hours and that the sample standard deviation of the lifetimes is 600 hours. Based on this information, answer the questions below. What are the null hypothesis (H.) and the alternative hypothesis (H,) that should be used for the test? Ho: u is ? H: u is ? In the context of this test, what is a Type I error? A Type I error is ? v the hypothesis that u is? when, in fact, u is ? Suppose that we decide to reject the null hypothesis. What sort of error might we be making? ?The average salary in this city is $47,400. Is the average less for single people? 70 randomly selected single people who were surveyed had an average salary of $45,379 and a standard deviation of $13,700. What can be concluded at the αα = 0.05 level of significance? For this study, we should use Correct The null and alternative hypotheses would be: H0:H0: Correct Correct H1:H1: Correct The test statistic = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is αα Based on this, we should the null hypothesis. Thus, the final conclusion is that ... The data suggest that the population mean is not significantly less than 47,400 at αα = 0.05, so there is statistically insignificant evidence to conclude that the population mean salary for singles is less than 47,400. The data suggest that the populaton mean is significantly less than 47,400 at αα…
