Computed p-value 4. p = 0.01 5. p = 0.20 6. p = 0.005 Level of Significance a a =0.05 a =0.05 a =0.01 Interpretation
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- The average expenditure per student (based on average daily attendance) for a certain school year was $10,337 with a population standard deviation of $1560. A survey for the next school year of 150 randomly selected students resulted in a sample mean of $10, 798. Find the Z statistics? Should the null hypothesis be rejected at alpha = .05 level of significance?The average expenditure per student (based on average daily attendance) for a certain school year was $10,337 with a population standard deviation of $1560. A survey for the next school year of 150 randomly selected students resulted in a sample mean of $10, 798. Find the P-value? Should the null hypothesis be rejected at alpha = .05 level of significance?A trucking firm is considering the installation of a new, low-restriction engine air filter for its long-haul trucks, but doesn’t want to make the switch unless the new filter can be shown to improve the fuel economy of these vehicles. A test is set up in which each of 10 trucks makes the same run twice once with the old filtration system and once with the new version. Given the sample results shown below, use the 0.05 level of significance in determining whether the new filtration system could be superior. Truck Number Current Filter (mpg) New Filter (mpg) 1 7.6 7.3 2 4.1 7.2 3 10.4 6.8 4 6.9 10.6 5 5.6 8.8 6 7.9 8.7 7 5.4 5.7 8 5.7 8.7 9 5.5 8.9 10 5.3 7.1 Do reject or fail to reject H0?
- Listed below are the heights (in inches) of fathers and their first sons. The dataare from a journal kept by Francis Galton. Use a 0.05 significance level to testthe claim that there is no difference in heights between fathers and their firstsons. (State the null and alternate hypothesis, the p-value method to make adecision, and write a conclusion based on the results.) Height of Father 72.0 66.0 69.0 70.0 70.0 70.0 70.0 75.0 68.265.0Height of Son 73.0 68.0 68.0 71.0 70.0 70.0 71.0 71.0 70.063.0For a special pre-New Year's Eve show, a radio station personality has invited a small panel of prominent local citizens to help demonstrate to listeners the adverse effect of alcohol on reaction time, thus drinking an alcohol increases reaction time. The reaction times (in seconds) before and after consuming four drinks are in Sheet 11. Test a hypothesis to see if the showman claim is supported by the given data (use alpha=0.01 significance level). Sheet 11 Subject Before After 1 0,32 0,39 2 0,39 0,44 3 0,36 0,49 4 0,41 0,53 5 0,37 0,46 6 0,35 0,39 7 0,37 0,49 8 0,37 0,52 9 0,32 0,43 10 0,39 0,45 11 0,39 0,39 12 0,41 0,49 13 0,32 0,48 14 0,38 0,48 15 0,34 0,4 16 0,35 0,52 Select one: a. The alternative hypothesis that the reaction time before consuming alcohol is less than after is not accepted as the p-value= 0.0000 is less than alpha=0.005 b. The Null hypothesis that the reaction time before consuming alcohol is less than after is rejected…The NAEP considers that a national average of 283 is an acceptable performance. Using α = .05, run a two-tail t-test for one sample to test Ho: µ=283 for the 2019 scores. Report the t-obt, df, and p-values. Would you reject the null hypothesis that the 2019 scores come from a population with average 283? If this is the case, does it come from a population from larger or smaller average?
- This chart shows the results of two random samples that measured the average number of minutes per charge for AA Lithium-ion (Li-ion) rechargeable batteries versus Nickel-Metal Hydride (NiMH) rechargeable batteries. Down below shows the hypothesis test using significance level (α) = 0.05 to determine if the true average number of minutes per charge for NiMH batteries is smaller than that for Li-ion batteries. 1. From the data given from the first graph below, what would be the correct p value? (the one tail or the two tail?) t-Test: Two-Sample Assuming Unequal Variances NiMH Li-ion Mean 89.35714 95 Variance 3.93956 59.75 Observations 14 17 Hypothesized Mean Difference 0 df 19 t Stat -2.89621 P(T<=t) one-tail 0.004628 t Critical one-tail 1.729133 P(T<=t) two-tail 0.009255 t Critical two-tail 2.093024 For the bottom graph: 1.. Find the point estimate (you can do this by subtracting Group 2…Five samples of a ferrous-type substance were used to determine if there is a difference between a laboratory chemical analysis and an X-ray fluorescence analysis of the iron content. Each sample was split into two subsamples and the two types of analysis were applied, with the accompanying results. Assuming that the populations are normal, test at the 0.05 level of significance whether the two methods of analysis give, on the average, the same result. Determine the test statistic t=?A sample of 20 third-grade students had a average of 54 on a math proficiency test, with a sample standard deviation of 11. Is there enough evidence to conclude that the third-grade sample mean differs significantly from the third-grade population mean of 67? Assume a 0.02 significance level. Use the Critical Value Method of Testing (this means NO P-Values!). will need to have -1. The null hypothesis, Ho 2. The alternative hypothesis, H1 3. The test statistic4. The type of test(left, right, two-tailed) and the p-value 5. The decision to accept Ho or reject Ho
- The Firearm Injury Surveillance Study (FISS) records whether the shooting arose out of a fight and the type of firearm used to cause the injury (here, handguns vs. rifles and shotguns). With an alpha of .01, conduct a five-step hypothesis test to determine if the variables are independent. Please fill out the chart provided and provide calculations. Fight Involved Handgun Long Gun Row Marginal Yes 158A 11B 169 No 309C 133D 442 Column Marginal 467 144 N = 611 Expected Frequency Count fe Calculations χ2obt Calculations:Listed in the data table are amounts of strontium-90 (in millibecquerels, or mBq, per gram of calcium) in a simple random sample of baby teeth obtained from residents in two cities. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Use a 0.100.10 significance level to test the claim that the mean amount of strontium-90 from city #1 residents is greater than the mean amount from city #2 residents.Two samples of sizes 25 and 20 are independently drawn from two normal populations, where the unknown population variances are assumed to be equal. The number of degrees of freedom of the equal-variances t-test statistic is ?