A biologist argues that a certain species of fish found in Lake Tanzania in Africa should be classified in a different genus than the currently accepted genus for the species. A simple random sample of standard lengths of fully grown fish from the species in Lake Tanzania is given below:{116.2, 100.3, 118, 152.8, 141.3, 132.8, 95.6, 111.5, 95.2, 95.7}The population of all such lengths is assumed to be close to normally distributed. Calculate the following interval estimates using this sample (note that only one of the methods below can be used for a single sample in practice). Round answers to one decimal place accuracy.A 99% confidence interval for the mean of all fully grown fish of this type is:[type your answer...] [type your answer...]A 99% prediction interval for the length of one fully grown fish of this type is:[type your answer...] [type your answer...]

Followings are the Explanation of the question Use the Excel and find the Sample Mean and Sample…

Answered: A biologist argues that a certain…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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Suppose that a certain chromosone defect occurs in 12% of all sheep. A random sample of 200 sheep is to be obtained and analysed for the chromosone defect. What is the mean of the sampling distribution of the sample proportion?
The U.S. Energy Information Administration claimed that U.S. residential customers used an average of 10,608 kilowatt hours (kWh) of electricity this year. A local power company believes that residents in their area use more electricity on average than EIA's reported average. To test their claim, the company chooses a random sample of 187 of their customers and calculates that these customers used an average of 10,737kWh of electricity last year. Assuming that the population standard deviation is 1220kWh, is there sufficient evidence to support the power company's claim at the 0.05 level of significance? Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.
The U.S. Energy Information Administration claimed that U.S. residential customers used an average of 10,476 kilowatt hours (kWh) of electricity this year. A local power company believes that residents in their area use more electricity on average than EIA's reported average. To test their claim, the company chooses a random sample of 153 of their customers and calculates that these customers used an average of 10,767 kWh of electricity last year. Assuming that the population standard deviation is 2478 kWh, is there sufficient evidence to support the power company's claim at the 0.05 level of significance? Step 3 of 3: Draw a conclusion and interpret the decision.
An expert reviews a sample of 10 scientific articles and records the following number of errors in each article: 0, 4, 2, 8, 2, 3, 1, 0, 5,7. Compute the mean, SS, variance, and standard deviation for this sample using the computational formula.
The U.S. Energy Information Administration claimed that U.S. residential customers used an average of 10,069 kilowatt hours (kWh) of electricity this year. A local power company believes that residents in their area use more electricity on average than EIA's reported average. To test their claim, the company chooses a random sample of 171 of their customers and calculates that these customers used an average of 10,461 kWh of electricity last year. Assuming that the population standard deviation is 2973 kWh, is there sufficient evidence to support the power company's claim at the 0.01 level of significance? Step 3 of 3: Draw a conclusion and interpret the decision.
The U.S. Energy Information Administration claimed that U.S. residential customers used an average of 10,941 kilowatt hours (kWh) of electricity this year. A local power company believes that residents in their area use more electricity on average than EIA's reported average. To test their claim, the company chooses a random sample of 115 of their customers and calculates that these customers used an average of 11,425 kWh of electricity last year. Assuming that the population standard deviation is 3217 kWh, is there sufficient evidence to support the power company's claim at the 0.02 level of significance? Step 3 of 3: Draw a conclusion and interpret the decision.
In a study comparing banks in two countries, a sample of 142 matched pairs of banks was formed. Each pair contained one bank from country x and one from country y. The pairings were made in such a way that the two members were as similar as possible in regard to such factors as size and age. The ratio of total loans outstanding to total assets was calculated for each of the banks. For this ratio, the sample mean difference (country x - country y) was 0.0579, and the sample standard deviation of the differences was 0.3086. Test, against a two-sided alternative, the null hypothesis that the two population means are equal. (Assume α = 0.01.) Click the icon to view the critical values of the Student's t distribution. Which of the following correctly states Ho and H₁ ? A. Ho: Mx Hy = 0 H₁: Hx-Hy >0 OC. Ho: Mx Hy ‡0 Hi HxHy=0 - O E. Ho: Mx Hy = 0 H₁: Hx-Hy <0 B. Ho: Mx Hy = 0 H₁: Hx-Hy #0 D. Ho: Hx-Hy = 0 H₁: Hx-Hy 20 OF. Ho: Mx Hy = = 0 H₁: Hx-Hy ≤0
The U.S. Energy Information Administration claimed that U.S. residential customers used an average of 10,069 kilowatt hours (kWh) of electricity this year. A local power company believes that residents in their area use more electricity on average than EIA's reported average. To test their claim, the company chooses a random sample of 171 of their customers and calculates that these customers used an average of 10,461 kWh of electricity last year. Assuming that the population standard deviation is 2973 kWh, is there sufficient evidence to support the power company's claim at the 0.01 level of significance? Step 2 of 3: Compute the value of the test statistic. Round your answer to two decimal places.
Compute the mean weight of a sample of seven chickens selected at random from a large poultry. the chickens weigh: 8,12,9,11.5,8.5,9.5,10( pounds)
In a study comparing banks in two countries, a sample of 142 matched pairs of banks was formed. Each pair contained one bank from country x and one from country y. The pairings were made in such a way that the two members were as similar as possible in regard to such factors as size and age. The ratio of total loans outstanding to total assets was calculated for each of the banks. For this ratio, the sample mean difference (country x - country y) was 0.0579, and the sample standard deviation of the differences was 0.3086. Test, against a two-sided alternative, the null hypothesis that the two population means are equal. (Assume α = 0.01.) Click the icon to view the critical values of the Student's t distribution. The test statistic is t = 2.236. (Round to three decimal places as needed.) The critical value(s) is(are) (Round to three decimal places as needed. Use a comma to separate answers as needed.)
A lawn mower company will produce 1,500 lawn mowers by 2020. In an effort to determine how much maintenance-free consumers will buy when they buy lawn mowers, the company decided to conduct a study over the past 1 year of these mowers. First, a random sample was taken by contacting 200 grass mowers. The company owner provides an 800 number and is asked to contact the company when the first repair is needed for the lawn mower. Owners who no longer use lawn mowers to cut their grass are either disqualified or not sampled. After 1 year, it turned out that 187 owners had reported for the first repair. However, the other 74 disqualified themselves. The average number per year until the first improvement was 5.3 times for samples deemed feasible. It is believed that the standard deviation is 1.28 per year. If the company wants to advertise the lawnmower and with a confidence level of 95%, what is the estimated annual mean value of repair-free lawn mower for this lawn mower? What is the…
A researcher believes that among the residents of a remote community, 30% have type O blood, 50% have type A blood, 10% have type B blood and 10% have type AB blood. She takes a random sample of 80 residents and finds that 20 had type O blood, 40 had type A blood, 12 had type B blood and 8 had type AB blood. Do the data follow the hypothesized distribution of blood type for this community? Test at a = .05. Round your answers to three decimal places, if necessary. Null hypothesis: Ho: Po = 0.3, PA = 0.5, PB = 0.1, PAB = 0.1 20 40 12 8 80 80 80 80 1 1 1 20 40 8 Ho: Po = Ho: Po = Alternative hypothesis: Ho: Po = PA = PB = PAB = Ha: po # Type O 1 Ha: Po PA PB = PAB = 4 20 › PA = PA = Type 0 20 24 40 80 80 Ha: Po 0.3, PA # 0.5, PB At least one of the proportions is different. Complete the table of observed counts: Type O , PA # 0.667 Type A PB = PB = 40 Type A 40 What is the test statistic? 2.667 1 ,PB # 0 , PAB = 1 12¹ PAB Complete the table of expected counts under the null hypothesis: 12…

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