The image shows a question concerning hypothesis testing with options for conclusions based on statistical significance. The question is structured as follows:**(c) Based on your answer to part (b), choose what can be concluded, at the 0.10 level of significance, about the claim made by the company.**1. **Option 1:** - Since the *p*-value is less than (or equal to) the level of significance, the null hypothesis is rejected. So, there is enough evidence to reject the claim that the mean number of minutes employees spend dealing with email is equal to 151.2. **Option 2:** - Since the *p*-value is less than (or equal to) the level of significance, the null hypothesis is not rejected. So, there is not enough evidence to reject the claim that the mean number of minutes employees spend dealing with email is equal to 151.3. **Option 3:** - Since the *p*-value is greater than the level of significance, the null hypothesis is rejected. So, there is enough evidence to reject the claim that the mean number of minutes employees spend dealing with email is equal to 151.4. **Option 4:** - Since the *p*-value is greater than the level of significance, the null hypothesis is not rejected. So, there is not enough evidence to reject the claim that the mean number of minutes employees spend dealing with email is equal to 151.There are also two navigation boxes with icons. One appears to be for exiting this section (an "X" icon), and another seems to be for returning or reviewing information (a "⟲" icon), suggesting an educational software interface context.Additionally, there is a graph above that shows a linear number line with markings from -3 to 3, which typically represents a standard normal distribution or a critical value region in hypothesis testing. However, there are no specific values or shaded areas provided to indicate critical regions or confidence intervals in this image. A very large company is interested in its employees' productivity. The company reports from its historical data that its employees spend a mean of 151 minutes per employee (on a typical day) dealing with email. To test this claim, an independent consultant chooses 21 employees at random and finds that those employees spend a sample mean of 160 minutes dealing with email, with a sample standard deviation of 21 minutes. Assume that the population of amounts of time employees spend dealing with email is approximately normally distributed.Complete the parts below to perform a hypothesis test to see if there is enough evidence, at the 0.10 level of significance, to reject the claim that μ, the mean number of minutes employees spend dealing with email, is equal to 151.(a) State the null hypothesis $H_0$ and the alternative hypothesis $H_1$ that you would use for the test.- Null hypothesis $H_0$: $\mu = 151$- Alternative hypothesis $H_1$: $\mu \ne 151$Diagram:The image contains a selection box to fill in the hypotheses with the following symbols:- $\mu$ (population mean)- $\bar{x}$ (sample mean)- $\neq, =, \leq, \geq$ (inequality symbols)- A reset button and an unknown (?) button are also present.

The sample size is 21. The sample mean is 160. The sample standard deviation is 21. We have to test…

Answered: A very large company is interested in…

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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An article in the San Jose Mercury News stated that students in the California state universitysystem take 4.25 years, on average, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of 64 students and obtain a sample mean of 4.5 with a sample standard deviation of 1.5. Do the data support your claim at the 1% level?
Average adult Americans are about one inch taller but nearly a whopping 25 pounds heavier than they were in 1960, according to a report from the Centers for Disease Control and Prevention (CDC). City A is considered one of America's healthiest cities. Is the weight gain since 1960 similar in city A? A sample of n=25 adults suggested a mean increase of 22 pounds with a standard deviation of 9.4 pounds. Is City A significantly different in terms of weight gain since 1960? Complete and interpret 5 step hypothesis testing by hand at a 5% level of significance using a Z-test. Show each step and your work.
A certain prescription medicine is supposed to contain an average of 250 parts per million (ppm) of a certain chemical. If the concentration is higher than this, the drug may cause harmful side effects; if it is lower, the drug may be ineffective. The manufacturer runs a check to see if the mean concentration in a large shipment conforms to the target level of 250 ppm or not. A simple random sample of 100 portions is tested, and the sample mean concentration is found to be 247 ppm. The sample concentration standard deviation is s = 12 ppm. What are the appropriate null and alternative hypotheses? Group of answer choices H 0: x̄ = 250 vs. H a: x̄ < 250 H 0: x̄ = 250 vs. H a: x̄ ≠ 250 H 0: x̄ = 250 vs. H a: x̄ > 250 H 0: μ = 250 vs. H a:μ < 250 H 0: μ = 250 vs. H a: μ ≠ 250 H 0: μ = 250 vs. H a: μ > 250
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A study was conducted for an aviation safety agency in which they surveyed thousands of passengers at airports across Europe. In this study, the "mass" for a passenger includes the weight of the passenger and all of the passenger's carry-on items, including infants without their own seats. For the 5,904 adult male passengers measured in summer, the sample mean and standard deviation for this mass were 88.8 kg and 15.9 kg, respectively. (a) Construct a 95% confidence interval (in kilograms) for the population mean mass for adult male passengers in summer. (Round your answers to three decimal places.) ___to__ kg (b) Write a sentence or two interpreting the confidence interval you found in part (a). With 95% confidence, we can estimate that the population mean mass of male passengers traveling in the summer is contained within the interval. We can estimate that 95% of the male passengers traveling in the summer in the sample have a mass that is contained within the interval.…
A parenting magazine reports that the average amount of wireless data used by teenagers each month is 5 Gb. For her science fair project, Ella sets out to prove the magazine wrong. She claims that the mean among teenagers in her area is less than reported. Ella collects information from a simple random sample of 25 teenagers at her high school, and calculates a mean of 4.7 Gb per month with a standard deviation of 0.9 Gb per month. Assume that the population distribution is approximately normal. Test Ella's claim at the 0.01 level of significance. Step 3 of 3: Draw a conclusion and interpret the decision. E Tables E Keypad Answer Keyboard Shortcuts We reject the null hypothesis and conclude that there is sufficient evidence at a 0.01 level of significance that the average amount of wireless data used by teenagers each month is less than 5 Gb. We fail to reject the null hypothesis and conclude that there is insufficient evidence at a 0.01 level of significance that the average amount of…
A parenting magazine reports that the average amount of wireless data used by teenagers each month is 10 Gb. For her science fair project, Ella sets out to prove the magazine wrong. She claims that the mean among teenagers in her area is less than reported. Ella collects information from a simple random sample of 16 teenagers at her high school, and calculates a mean of 9.1 Gb per month with a standard deviation of 1.3 Gb per month. Assume that the population distribution is approximately normal. Test Ella's claim at the 0.05 level of significance. Step 3 of 3: Draw a conclusion and interpret the decision.

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