sample Standard deviation = 42 sample size of Group 1 = 10   Disney World (Group 2): Average number of people per day = 520.26 Sample Standard deviation = 35 sample size of Group 2 = 22   Conduct a hypothesis test based on the research question presented.  Using an alpha of .1 Reject the null hypothesis Fail to reject the null hypothesis Reject the alternative hypothesis Fail to reject the alternative hypothesis

2. A professional racing team is holding tryouts at its test track. The team requires its drivers to have an average lap time of at most 133 seconds at its test track. The team will use a one sample z-test of significance to determine if they should hire a new driver.   An amateur driver completes 10 timed laps of the test track with an average lap time of 140 seconds. If the standard deviation of all lap times at this track is 4.5 seconds, what is the value of the test statistic? Round your answer to two decimal places. Answer value    Please show work!!!

12) A state fish hatchery raises trout for stocking streams and lakes. The size of the fish at release time can be controlled to a fair degree by varying the rate of feeding. The target is a mean of 10 ounces; if the fish are too small, those who catch the fish aren’t happy. A random sample of 75 fish were weighed at time of release and it was determined that the mean was 9.66 ounces with a standard deviation of 0.86 ounces. Test to determine if the fish being released have a population mean less than 10 ounces at the 0.01 significance level.

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Group 1. Mean= 24.08                    Group 2. Mean= 26.5            STD Dev= 4.42                                STD Dev= 5.2            Sample Size=16                            Sample Size=16   What is tcomputed =

Low-fat or low-carb? Are low-fat diets more effective for weight loss? A sample of 46 subjects went on a low carbohydrate diet for six months. At the end of that time, the sample mean weight loss was 4.6 kilograms with a sample standard deviation of 6.28 kilograms. A second sample of 49 subjects went on a low-fat diet. Their sample mean weight loss was 2.3 kilograms with a standard deviation of 4.66 kilograms. Can you conclude that the mean weight loss differs between the two diets? Let μ1 denote the mean weight lost on the low-carb diet and μ2 denote the mean weight lost on the low-fat diet. Use the =α0.01level and the P-value method.   Compute the test statistic. Round the answer to three decimal places. t =

Perform an analysis of variance (ANOVA) to determine if the differences in the percentage butterfat of the different breeds of cows is statistically significant. Guernsey Jersey Holstein 4.51 4.59 3.43 5.26 6.67 3.68 5.81 5.39 3.72 5.14 4.38 3.76 4.55 4.99 4.43 4.86 5.49 3.84 4.69 5.43 3.54 4.12 5.02 4.18 5.44 5.16 3.4 4.69 5.15 3.77 5.24 5.76 4.03 4.62 5.34 3.63 4.61 5.11 3.37 5.02 4.75 3.54 4.67 5.07 3.64 4.59 4.42 3.98 4.92 6.13   5.58 4.6   5.6     5.83   1. Compute the treatment sum of squares (SSTr) and the error sum of squares (SSE) and use them to complete the following ANOVA table. (Round your answers to 4 decimal places). Source S.S. df M.S. F Treatment         Error         Total           2. Compute the p-value. (Round your answer to 4 decimal places.)p-value =

Community college instructors' salaries in one state are very low, so low that educators in that state regularly complain about their compensation. The national mean is $57,636$⁢57,636, but instructors from Mississippi claim that the mean in their state is significantly lower. They survey a simple random sample of 3737 colleges in the state and calculate a mean salary of $52,534$⁢52,534 with a standard deviation of $15,791$⁢15,791. Test the instructors' claim at the 0.050.05 level of significance.   Step 3 of 3 :  Draw a conclusion and interpret the decision.

Jess2.4 employees leaving per month, with a standard deviation of 0.5. She decided to see if using a new benefits package would affect her average. She used the new benefits package for 13 months. For those 13 months, Jessica got an average of 2.6 employees leaving. She wants to know if the new benefits package caused her to get a different number of employees leaving per month.  She conducts a one-mean hypothesis at the 5% significance level, to see if the new benefits package caused her to get a different number of employees leaving per month.   Which answer choice shows the correct null and alternative hypotheses for this test? Select the correct answer below:   H0:μ=2.4; Ha:μ≠2.4, which is a two-tailed test.   H0:μ=2.4; Ha:μ<2.4, which is a left-tailed test.   H0:μ=2.4; Ha:μ>2.4, which is a right-tailed test.   H0:μ=2.6; Ha:μ≠2.6, which is a left-tailed test.