The mean number of English courses taken in a two-year time period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from 26 males and 20 females. The males took an average of 3.7 English courses with a standard deviation of 0.3. The females took an average of 3.6 English courses with a standard deviation of 0.6. Conduct a hypothesis test at the 1% level of significance to determine whether the means are statistically the same. Step 1: State the null and alternative hypotheses. Ho: 12V Ha: 1 - ₂0 - 0 (So we will be performing a two-tailed The differences of sample means are t-distributed and distribution standard deviation 10.3 Question Help: D Post to forum Submit Part test.) Part 2 of 4 Step 2: Assuming the null hypothesis is true, determine the features of the distribution of the differences of sample means. Part 1 of 4 with distribution mean o

An economist was curious if the mean yearly amount of seafood consumed in Sweden is significantly different than the mean yearly consumption in the United States. They randomly selected 125 citizens in Sweden and determined their sample mean is 12 kg with a standard deviation of 5.2 kg. The 140 randomly selected U.S. citizens had a sample mean of 9.7 kg with a standard deviation of 5.8 kg. Is this sufficient evidence to say that the mean consumption is different? Which are the correct hypothesis statements for this situation? Hoμs Hus HAμs=Pus H。 ¦ µ¸ = Pus HAμs Hus Hoµs=Pus HAHs>Hus Hop=0.10 ΗΑ : μ > 0. 10

The mean number of English courses taken in a two-year time period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from 24 males and 26 females. The males took an average of 4.2 English courses with a standard deviation of 1.1. The females took an average of 3.9 English courses with a standard deviation of 0.7. Conduct a hypothesis test at the 4% level of significance to determine whether the means are statistically the same. Step 1: State the null and alternative hypotheses. Ho: H1 = V H.: H1 – µ2 # v test.) (So we will be performing a two-tailed

Researchers from a certain country were interested in how characteristics of the spleen of residents in their tropical environment compare to those found elsewhere in the world. The researchers randomly sampled 93 males and 107 females in their country. The mean and standard deviation of the spleen lengths for the males were 10.8 cm and 0.9 cm, respectively, and those for the females were 10.2 cm and 0.7 cm, respectively. At the 5% significance level, do the data provide sufficient evidence to conclude that a difference exists in the mean spleen lengths of males and females in the country? Click here to view page 1 of the table of critical values of t. Click here to view page 2 of the table of critical values of t. O E. Ho: H1 H₂ Hai Hg = H2 Compute the test statistic. =(Round to two decimal places as needed.) Determine the critical value(s). t= OF. Ho: H1 H₂ Ha: H1 H₂ 0 (Round to three decimal places as needed. Use a comma to separate answers as needed.) What is the conclusion of the…

A custodian wishes to compare two competing floor waxes to decide which one is best. He believes that the mean of WaxWin is less than the mean of WaxCo. In a random sample of 18 floors of WaxWin and 10 of WaxCo. WaxWin had a mean lifetime of 22.1 with a standard deviation of 5.6 and WaxCo had a mean lifetime of 28.5 with a standard deviation of 7.1. Perform a hypothesis test using a significance level of 0.05 to help him decide. Let WaxWin be sample 1 and WaxCo be sample 2 The correct hypotheses are: O Ho: H1 µ2(claim) O Ho: µ1 2 H2 HA: H1 < µ2(claim) O Ho: µ1 = µ2 HA: H1 7 M2(claim) Since the level of significance is 0.05 the critical value is -1.751 The test statistic is: (round to 3 places) The p-value is: (round to 3 places) The decision can be made to: O reject Ho O do not reject Ho The final conclusion is that: O There is enough evidence to reject the claim that the mean of WaxWin is less than the mean of WaxCo. O There is not enough evidence to reject the claim that the mean of…

1) A study compared the weight loss of people on a low-fat diet versus people on a low-carb diet. In a sample of 100 obese people on a low-fat diet the sample mean weight loss as 7.6 pounds with a standard deviation of 3.2 pounds. In another sample of 120 obese people on a low-carb diet the sample mean weight loss was 6.7 pounds with a standard deviation of 3.9 pounds.   a) Test the hypothesis, at the 5% level of significance, that there is a difference in the mean weight loss of using the 2 different diets. b) What is the p-value?

An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 3939 type K batteries and a sample of 5757 type Q batteries. The mean voltage is measured as 8.558.55 for the type K batteries with a standard deviation of 0.6830.683, and the mean voltage is 8.828.82 for type Q batteries with a standard deviation of 0.7910.791. Conduct a hypothesis test for the conjecture that the mean voltage for these two types of batteries is different. Let μ1μ1 be the true mean voltage for type K batteries and μ2μ2 be the true mean voltage for type Q batteries. Use a 0.020.02 level of significance. Step 3 of 4 :   Determine the decision rule for rejecting the null hypothesis H0H0. Round the numerical portion of your answer to two decimal places

A study is conducted to find the mean life of a certain manufacturer's car batteries. A random sample of 34 batteries is selected and a mean of 47.5 months with a standard deviation of 7.3 months is found. Find a 95% confidence interval for the mean battery life.

Bone mineral density (BMD) is a measure of bone strength. Studies show that BMD declines after age 45. The impact of exercise may increase BMD. A random sample of 59 women between the ages of 41 and 45 with no major health problems were studied. The women were classified into one of two groups based upon their level of exercise activity: walking women and sedentary women. The 39 women who walked regularly had a mean BMD of 5.96 with a standard deviation of 1.22. The 20 women who are sedentary had a mean BMD of 4.41 with a standard deviation of 1.02. Which of the following inference procedures could be used to estimate the difference in the mean BMD for these two types of women

In September a biological research team caught, weighed, and released a random sample of 24 chipmunks in Rocky Mountain National Park. The mean of the sample weights was x-bar = 8.7 oz. with sample standard deviation s = 1.4 oz. Find a 90% confidence interval for the mean September weight of all chipmunks in Rocky Mountain National Park.