A researcher claims that the stomachs of blue crabs from Location A contain more fish than the stomachs of blue crabs from Location B. The stomach contents of asample of 14 blue crabs from Location A contain a mean of 190 milligrams of fish and a standard deviation of 36 milligrams. The stomach contents of a sample of11 blue crabs from Location B contain a mean of 189 milligrams of fish and a standard deviation of 43 milligrams. At a 0.05, can you support the researcher'sclaim? Assume the population variances are equal. Complete parts (a) through (d) below.(a) Identify the null and alternative hypotheses. Choose the correct answer below.CA. Hồi đây - Hày 20H₂: P1 - H₂ <0OC. Ho: 11-1₂ 50H₂:4₁-4₂0OB. Ho: H₁-H₂=0На 1-H2*0OD. Ho: H₁-H₂ <0Ha: 11-H2=0

The given researcher claims is that the stomachs of blue crabs from Location A contains more fish…

Answered: A researcher claims that the stomachs…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A researcher claims that the stomachs of blue crabs from Location A contain more fish than the stomachs of blue crabs from Location B. The stomach contents of a sample of 14 blue crabs from Location A contain a mean of 194 milligrams of fish and a standard deviation of 37 milligrams. The stomach contents of a sample of 8 blue crabs from Location B contain a mean of 184 milligrams of fish and a standard deviation of 44 milligrams. At a = 0.05, can you support the researcher's claim? Assume the population variances are equal. Complete parts (a) through (d) below. (a) Identify the null and alternative hypotheses. Choose the correct answer below. ОА. Но H1-H2 0 OD. Ho: H₁-12₂=0 Ha H1 H₂0 enough evidence at the 5% level of significance to support the researcher's claim.
A new small business wants to know if its current radio advertising is effective. The owners decide to look at the mean number of customers who make a purchase in the store on days immediately following days when the radio ads are played as compared to the mean for those days following days when no radio advertisements are played. They found that for 10 days following no advertisements, the mean was 18.3 purchasing customers with a standard deviation of 1.8 customers. On 7 days following advertising, the mean was 19.4 purchasing customers with a standard deviation of 1.6 customers. Test the claim, at the 0.02 level, that the mean number of customers who make a purchase in the store is lower for days following no advertising compared to days following advertising. Assume that both populations are approximately normal and that the population variances are equal. Let days following no advertisements be Population 1 and let days following advertising be Population 2. Step 3 of 3: Draw a…
A researcher claims that the stomachs of blue crabs from Location A contain more fish than the stomachs of blue crabs from Location B. The stomach contents of a sample of 13 blue crabs from Location A contain a mean of 191 milligrams of fish and a standard deviation of 37 milligrams. The stomach contents of a sample of 7 blue crabs from Location B contain a mean of 185 milligrams of fish and a standard deviation of 41 milligrams. At α=0.10, can you support the researcher's claim? Assume the population variances are equal and that both samples are from normal populations. Complete parts (a) through (d) below. (a) Identify the null and alternative hypotheses. Choose the correct answer below. A. H0: μ1−μ2=0 Ha: μ1−μ2>0 B. H0: μ1−μ2<0 Ha: μ1−μ2=0 C. H0: μ1−μ2=0 Ha: μ1−μ2≠0 D. H0: μ1−μ2=0 Ha: μ1−μ2<0 (b) Calculate the test statistic t=? (Round to three decimal places as needed.) (c) Calculate the…
A car company claims that its new SUV gets better gas mileage than its competitor's SUV. A random sample of 35 of its SUVs has a mean gas mileage of 16.7 miles per gallon (mpg). The population standard deviation is known to be 0.9 mpg. A random sample of 42 competitor's SUVS has a mean gas mileage of 16.4 mpg. The population standard deviation for the competitor is known to be 0.6 mpg. Test the company's claim at the 0.02 level of significance. Let the car company's SUVs be Population 1 and let the competitor's SUVS be Population 2. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. Họ :-1 = M2 Ha М2
The heights of 10 year old American children are normally distributed with a mean height of 4 feet and 5 inches (or 53 inches) and a standard deviation of 4.5 inches.In order to be allowed to go on a particular kids ride at the amusement park, children must be at least 44.9 inches tall and no more than 59.75 inches tall.What percent of 10 year old American children cannot go on the ride?
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A new small business wants to know if its current radio advertising is effective. The owners decide to look at the mean number of customers who make a purchase in the store on days immediately following days when the radio ads are played as compared to the mean for those days following days when no radio advertisements are played. They found that for 13 days following no advertisements, the mean was 23.9 purchasing customers with a standard deviation of 1.9 customers. On 6 days following advertising, the mean was 24.7 purchasing customers with a standard deviation of 1.6 customers. Test the claim, at the 0.01 level, that the mean number of customers who make a purchase in the store is lower for days following no advertising compared to days following advertising. Assume that both populations are approximately normal and that the population variances are equal. Let days following no advertisements be Population 1 and let days following advertising be Population 2. Step 3 of 3: Draw a…
Aresearcher claims that the stomachs of blue crabs from Location A contain more fish than the stomachs of blue crabs from Location B. The stomach contents of a sample of 14 blue crabs from Location A contain a mean of 194 milligrams of fish and a standard deviation of 39 milligrams. The stomach contents of a sample of 8 blue crabs from Location B contain a mean of 187 milligrams of fish and a standard deviation of 44 milligrams. Al a 0 10, can you support the researcher's claim? Assume the population variances are equal. Complete parts (a) through (d) below. (a) Identify the null and alternative hypotheses. Choose the correct answer below. O A. Ho: H-220 O B. Ho H-2 0 (b) Find the standardized test statistic for u - t=0.381 (Round to three decimal places as needed) Enter your answer in the answer box and then click Check Answer parts remaining Clear All Check Answer leste ppt_08_03.pptx Open file Show all P Type here to search 235 PM 2/21/2021 ts fho sk hz home end insert delet esc & @…
A researcher claims that the stomachs of blue crabs from Location A contain more fish than the stomachs of blue crabs from Location B. The stomach contents of a sample of 13 blue crabs from Location A contain a mean of 191 milligrams of fish and a standard deviation of 38 milligrams. The stomach contents of a sample of 8 blue crabs from Location B contain a mean of 189 milligrams of fish and a standard deviation of 44 milligrams. At a = 0.10, can you support the researcher's claim? Assume the population variances are equal. Complete parts (a) through (d) below. t= 0.110 (Round to three decimal places as needed.) (c) Calculate the P-value. P=|| (Round to four decimal places as needed.)
A study is done to determine which of two soft drinks has more sugar. There are 25 cans of Beverage A in the researchers' sample and 15 cans of Beverage B. The mean amount of sugar in Beverage A is 34.7 grams per can with a sample standard deviation of 0.7 grams. The mean amount of sugar in Beverage B is 33.9 grams with a standard deviation of 0.6 grams. The researchers believe that Beverage A has less sugar than Beverage B, on average. Both populations are normally distributed. Conduct a hypothesis test using a 2% level of significance. Step 1: State the null and alternative hypotheses. Ho:µA – µB 2V Ho:HA - HB V test.) (So we will be performing a left-tailed forum
The heights of adult men in America are normally distributed, with a mean of 69.5 inches and a standard deviation of 2.61 inches. The heights of adult women in America are also normally distributed, but with a mean of 64.2 inches and a standard deviation of 2.54 inches.

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