QUESTION AL A leading E-commerce company conducts a study to observe the weekend sales per customer over 50 years old is RM 5000. Based on the sample of 26 customers who were randomly selected, statistics show the mean spent for weekend sales is only RM 4200. The standard deviation of the sample is known to be RM 65.50 with a 1% level of significance. Compute the value of test statistic. (b) Explain why t-distribution is used for this hypothesis testing.
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- The Insure.com website reports that the mean annual premium for automobile insurance in the United States was $1,503 in March 2014. Being from Pennsylvania at that time, you believed automobile insurance was cheaper there and decided to develop statistical support for your opinion. A sample of 25 automobile insurance policies from the state of Pennsylvania showed a mean annual premium of $1,430 with a standard deviation of s = $160. (a) Develop a hypothesis test that can be used to determine whether the mean annual premium in Pennsylvania is lower than the national mean annual O Ho: u = 1,503 H: u # 1,503 premium. o Ho: H 1,503 O Ho: H > 1,503 Ha:uS 1,503 (b) What is a point estimate in dollars of the difference between the mean annual premium in Pennsylvania and the national mean? (Use the mean annual premium in Pennsylvania minus the national mean.)Is there any difference in the variability in golf scores for players on a women's professional golf tour and players on a men's professional golf tour? A sample of 20 tournament scores from events in a tour for women showed a standard deviation of 2.4628 strokes, and a sample of 30 tournament scores from events in a tour for men showed a standard deviation of 2.2173. Conduct a hypothesis test for equal population variances to determine if there is any statistically significant difference in the variability of golf scores for male and female professional golfers. Use a = 0.10. State the null and alternative hypotheses. 2 H_: 2 Ho H₂01 H: #02 2 2 02 2 022 202 Ho: 1 H₂ 2 1 02 Find the value of the test statistic. (Round your answer to two decimal places.) Find the p-value. (Round your answer to four decimal places.) p-value = State your conclusion. O Reject Ho. We cannot conclude that there is a difference in the variability of golf scores for male and female professional golfers. Do not…A study by the department of education of a certain state was trying to determine the mean SAT scores of the graduating high school seniors. The study examined the scores of a random sample of 162 graduating seniors and found the mean score to be 526 with a standard deviation of 87. Use the normal distribution/empirical rule to estimate a 95% confidence interval for the mean, rounding all values to the nearest tenth
- During a recent year, the mean ACT score for all college bound students had a mean of 24.65 and a standard deviation of 4.01. The director of Stanley Kaplan is interested in knowing whether taking their course causes a difference on ACT scores. She takes a random sample of 444 seniors and finds their mean to be 25.44. Test the null hypothesis at the .05 level of significance.A shareholders' group is lodging a protest against your company. The shareholders group claimed that the mean tenure for a chief exective office (CEO) was different 10 years. A survey of 75 companies reported in The Wall Street Journal found a sample mean tenure of 8.1 years for CEOs with a standard deviation of s= 4.4 years (The Wall Street Journal, January 2, 2007). You don't know the population standard deviation but can assume it is normally distributed.You want to formulate and test a hypothesis that can be used to challenge the validity of the claim made by the group, at a significance level of α=0.001. Your hypotheses are: Ho:μ=10 Ha:μ≠10What is the test statistic for this sample? test statistic = (Report answer accurate to 3 decimal places.)What is the p-value for this sample? p-value = (Report answer accurate to 4 decimal places.)The p-value is... less than (or equal to) α greater than α This test statistic leads to a decision to... reject the null accept the…According to a city's estimate, people on average arrive 1.5 hours early for domestic flights. The population standard deviation is known to be 0.5 hours. A researcher wanted to check if this is true so he took a random sample of 50 people taking domestic flights and found the mean time to be 2.0 hours early. At the 1% significance level, can you conclude that the amount of time people are early for domestic flights is more than what the city claims. Will you reject or not reject the null hypothesis and what is your conclusion in words? O Reject the null hypothesis; the city's claim is true. O Do not reject the null hypothesis; the city's claim is false and people arrive earlier than the city claims. O Reject the null hypothesis; the city's claim is false and people arrive earlier than the city claims. O Do not reject the null hypothesis; the city's claim is true.
- Is there any difference in the variability in golf scores for players on a women's professional golf tour and players on a men's professional golf tour? A sample of 20 tournament scores from events in a tour for women showed a standard deviation of 2.4638 strokes, and a sample of 30 tournament scores from events in a tour for men showed a standard deviation of 2.2121. Conduct a hypothesis test for equal population variances to determine if there is any statistically significant difference in the variability of golf scores for male and female professional golfers. Use a = 0.10. State the null and alternative hypotheses. 2 O Ho: o 2702 H: 01 Ho: 01 02 キ02 2ァ022 して 2. キ02 2502 2 %3D Find the value of the test statistic. (Round your answer to two decimal places.) Find the p-value. (Round your answer to four decimal places.) p-value %3D State your conclusion. O Reject Ho: We can conclude that there is a difference in the variability of golf scores for male and female professional golfers. O…The Insure.com website reports that the mean annual premium for automobile insurance in the United States was $1,503 in March 2014. Being from Pennsylvania at that time, you believed automobile insurance was cheaper there and decided to develop statistical support for your opinion. A sample of 25 automobile insurance policies from the state of Pennsylvania showed a mean annual premium of $1,425 with a standard deviation of s = $165. (a) Develop a hypothesis test that can be used to determine whether the mean annual premium in Pennsylvania is lower than the national mean annual premium. Ho: μ = 1,503 H₂: μ = 1,503 Ho: μ> 1,503 H₂:μ ≤ 1,503 Ho: μ 1,503 (b) What is a point estimate in dollars of the difference between the mean annual premium in Pennsylvania and the national mean? (Use the mean annual premium in Pennsylvania minus the national mean.) $ (c) At a = 0.05, test for a significant difference. Find the value of the test statistic. (Round your answer to three decimal places.) Find…Independent random samples of patients who had received knee and hip replacement were asked to assess the quality of service on a scale from 1 (low) to 7 (high). For a sample of 83 knee patients, the mean rating was 6.543 and the sample standard deviation was 0.649. For a sample of 54 hip patients, the mean rating was 6.733 and the sample standard deviation was 0.425. Test, against a two-sided alternative, the null hypothesis that the population mean ratings for these two types of patients are the same.
- 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 womenDuring 2008, college work-study students earned a mean of $1478. Assume that a sample consisting of 25 of the work-study students at a large university was found to have earned a mean of $1503 during that year, with a standard deviation of $210. The null and alternative hypotheses are formulate to suggest that the average earnings of this university’s work-study students were significantly higher than the national mean. At 5% level of significance, do reject or fail to reject H0? Fail to reject H0 because t-statistic is less than 1.711 Fail to reject H0 because t- statistic is greater than 1.711 Fail to reject H0 because t- statistic is greater than -1.711 Fail to reject H0 because t- statistic is less than -1.711The average undergraduate cost for tuition, fees, and room and board for two-year institutions last year was $14,124. The following year, a random sample of 20 two-year institutions had a mean of $17,124 and a standard deviation of $3700. Is there sufficient evidence at the alpha level of 0.05 to conclude that the mean cost has increased. Show 6 steps of hypothesis