Question

1. You measure 42 textbooks' weights, and find they have a mean weight of 47 ounces. Assume the population standard deviation is 3.5 ounces. Based on this, construct a 90% confidence interval for the true population mean textbook weight.

Give your answers as decimals, to two places

2.If n=16, ¯xx¯(x-bar)=43, and s=13, construct a confidence interval at a 99% confidence level. Assume the data came from a normally distributed population.

Give your answers to one decimal place.

3.SAT scores are distributed with a mean of 1,500 and a standard deviation of 300. You are interested in estimating the average SAT score of first year students at your college. If you would like to limit the margin of error of your 95% confidence interval to 25 points, how many students should you sample?

4.

A student was asked to find a 95% confidence interval for widget width using data from a random sample of size n = 16. Which of the following is a correct interpretation of the interval 10.5 < μ < 24.8?

Check all that are correct.

• There is a 95% chance that the mean of the population is between 10.5 and 24.8.
• The mean width of all widgets is between 10.5 and 24.8, 95% of the time. We know this is true because the mean of our sample is between 10.5 and 24.8.
• With 95% confidence, the mean width of all widgets is between 10.5 and 24.8.
• With 95% confidence, the mean width of a randomly selected widget will be between 10.5 and 24.8.
• There is a 95% chance that the mean of a sample of 16 widgets will be between 10.5 and 24.8.   

  5.You are a researcher studying the lifespan of a certain species of bacteria. A preliminary sample of 30 bacteria reveals a sample mean of ¯x=76x¯=76 hours with a standard deviation of s=6.2s=6.2 hours. You would like to estimate the mean lifespan for this species of bacteria to within a margin of error of 0.8 hours at a 99% level of confidence.

  What sample size should you gather to achieve a 0.8 hour margin of error? Round your answer up to the nearest whole number.

  n =   bacteria

• 6.Express the confidence interval (26.7%,37.5%)(26.7%,37.5%) in the form of ˆp±MEp^±ME.
  

• 7.

  A political candidate has asked you to conduct a poll to determine what percentage of people support her.
  
  If the candidate only wants a 10% margin of error at a 99% confidence level, what size of sample is needed?
  
  Give your answer in whole people.

  8.

  A student was asked to find a 99% confidence interval for the proportion of students who take notes using data from a random sample of size n = 86. Which of the following is a correct interpretation of the interval 0.14 < p < 0.28?

  Check all that are correct.

• With 99% confidence, the proportion of all students who take notes is between 0.14 and 0.28.
• There is a 99% chance that the proportion of notetakers in a sample of 86 students will be between 0.14 and 0.28.
• With 99% confidence, a randomly selected student takes notes in a proportion of their classes that is between 0.14 and 0.28.
• The proprtion of all students who take notes is between 0.14 and 0.28, 99% of the time.
• There is a 99% chance that the proportion of the population is between 0.14 and 0.28.
• 9.The mean weight of female nurses working at a local hospital is at most 130 lbs. Express the null and alternative hypotheses in symbolic form for this claim.
  
  H0:μH0:μ
  
  Ha:μHa:μ
  
  Use the following codes to enter the following symbols:
       ≥≥ enter >=
       ≤≤ enter <=
       ≠≠ enter !=
• 10.

  Food inspectors inspect samples of food products to see if they are safe. This can be thought of as a hypothesis test with the following hypotheses.

  H₀: the food is safe

  H_a: the food is not safe

  The following is an example of what type of error?

  The sample suggests that the food is safe, but it actually is not safe.

• type I
• type II
• not an error

11. You are performing a left-tailed test.

If α=.02α=.02, find the critical value, to three decimal places.

z_α =

12.

It is believed that nearsightedness affects about 8% of all children. In a random sample of 194 children, 21 are nearsighted.


(a) Construct hypotheses appropriate for the following question: do these data provide evidence that the 8% value is inaccurate?

• Ho: p = .08
  Ha: p ≠ .08
• Ho: p = .08
  Ha: p > .08
• Ho: p = .08
  Ha: p < .08

(b) What proportion of children in this sample are nearsighted?
(round to four decimal places)


(c) Given that the standard error of the sample proportion is 0.0195 and the point estimate follows a nearly normal distribution, calculate the test statistic (use the Z-statistic).
Z =  (please round to two decimal places)


(d) What is the p-value for this hypothesis test?
p =  (please round to four decimal places)


(e) What is the conclusion of the hypothesis test?

• Since p ≥ α we accept the null hypothesis
• Since p ≥ α we do not have enough evidence to reject the null hypothesis
• Since p<α we reject the null hypothesis and accept the alternative
• Since p ≥ α we reject the null hypothesis and accept the alternative
• Since p<α we fail to reject the null hypothesis

13.Test the claim that the mean GPA of night students is smaller than 2.6 at the 0.025 significance level.

The null and alternative hypothesis would be:

H0:μ=2.6H0:μ=2.6
H1:μ≠2.6H1:μ≠2.6

H0:p≥0.65H0:p≥0.65
H1:p<0.65H1:p<0.65

H0:μ≤2.6H0:μ≤2.6
H1:μ>2.6H1:μ>2.6

H0:μ≥2.6H0:μ≥2.6
H1:μ<2.6H1:μ<2.6

H0:p≤0.65H0:p≤0.65
H1:p>0.65H1:p>0.65

H0:p=0.65H0:p=0.65
H1:p≠0.65H1:p≠0.65

The test is:

two-tailed

right-tailed

left-tailed

Based on a sample of 75 people, the sample mean GPA was 2.58 with a standard deviation of 0.05

The p-value is:  (to 2 decimals)

Based on this we:

• Fail to reject the null hypothesis
• Reject the null hypothesis

14.Test the claim that the proportion of people who own cats is significantly different than 30% at the 0.02 significance level.

The null and alternative hypothesis would be:

H0:μ≥0.3H0:μ≥0.3
Ha:μ<0.3Ha:μ<0.3

H0:p≥0.3H0:p≥0.3
Ha:p<0.3Ha:p<0.3

H0:μ≤0.3H0:μ≤0.3
Ha:μ>0.3Ha:μ>0.3

H0:p≤0.3H0:p≤0.3
Ha:p>0.3Ha:p>0.3

H0:μ=0.3H0:μ=0.3
Ha:μ≠0.3Ha:μ≠0.3

H0:p=0.3H0:p=0.3
Ha:p≠0.3Ha:p≠0.3

The test is:

left-tailed

two-tailed

right-tailed

Based on a sample of 400 people, 23% owned cats

The p-value is:  (to 2 decimals)

Based on this we:

• Fail to reject the null hypothesis
• Reject the null hypothesis

15. You wish to test the following claim (HaHa) at a significance level of α=0.005α=0.005.

      Ho:μ=87.7Ho:μ=87.7
      Ha:μ>87.7Ha:μ>87.7

You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size n=10n=10 with mean M=103.8M=103.8 and a standard deviation of SD=15.4SD=15.4.

What is the p-value for this sample? (Report answer accurate to four decimal places.)
p-value =

The p-value is...

• less than (or equal to) αα
• greater than αα

This p-value leads to a decision to...

• reject the null
• accept the null
• fail to reject the null

As such, the final conclusion is that...

• There is sufficient evidence to warrant rejection of the claim that the population mean is greater than 87.7.
• There is not sufficient evidence to warrant rejection of the claim that the population mean is greater than 87.7.
• The sample data support the claim that the population mean is greater than 87.7.
• There is not sufficient sample evidence to support the claim that the population mean is greater than 87.7

Answer 1

As per Chegg guidelines ask each question separately