Question

Question 5 1 pts The mean result from some experiment with 50 observations was 1.5 and the standard deviation was 1. Let deno
0 0
Add a comment Improve this question Transcribed image text
Answer #1

Ha : \mu > 1

Hence a right tailed test.

T-Mo (1.5 -1) 0.5 Test Statistic : tstat =- /v 1/500 12=3,5361

Degrees of freedom = n-1 =50-1 = 49

For right tailed test :

p-value = P(+49 > tstat) = P(+49 > 3.5)

Ans : The p-value is

Plt49 > 3.5)

Add a comment
Know the answer?
Add Answer to:
Question 5 1 pts The mean result from some experiment with 50 observations was 1.5 and...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • The mean result from some experiment with 50 observations was 1.5 and the standard deviation was...

    The mean result from some experiment with 50 observations was 1.5 and the standard deviation was 1. Let μ denote the true mean result. We wish to to test H 0 : μ = 1 vs H a : μ ≠ 1 . The p-value is Group of answer choices 2×P(t49≤3.5) 2×P(t49≥|3.5|) 2×P(t49≤−3.5)

  • The mean result of some observations was 50. Let μ denote the true mean result. We...

    The mean result of some observations was 50. Let μ denote the true mean result. We wish to test the hypothesis H 0 : μ = 40 vs H a : μ < 40 . Without doing any calculations, the p-value and test conclusion are most likely Group of answer choices Close to 0 so reject the null hypothesis Close to 1 so fail to reject the null hypothesis Close to 0 so fail to reject the null hypothesis Close...

  • Of the sample of 72 observations for sparrows that survived, the mean humerus length was 0.74...

    Of the sample of 72 observations for sparrows that survived, the mean humerus length was 0.74 in, and the standard deviation was 0.02 in. And of the sample of 64 observations for sparrows that died, the mean humerus length was 0.73 in, and the standard deviation was 0.03 in. Let μs denote the mean for the population distribution of humerus length for sparrows that would survive, and let μd denote the mean for the population distribution of humerus length for...

  • A sample of 44 observations is selected from a normal population. The sample mean is 46,...

    A sample of 44 observations is selected from a normal population. The sample mean is 46, and the population standard deviation is 7. Conduct the following test of hypothesis using the 0.01 significance level. Ho: M = 50 Hu 50 a. Is this a one- or two-tailed test? One-tailed test Two-tailed test b. What is the decision rule? Reject Ho if -2.576 <z<2.576 Reject Ho if z<-2.576 or z> 2.576 c. What is the value of the test statistic? (Negative...

  • Question 22 1 pts A survey conducted in a small business yielded the result shown in...

    Question 22 1 pts A survey conducted in a small business yielded the result shown in the table. Men Women Total Plan to 41 22 vote Do not plan 34 24 to vote Total Test the independence. Which of the following is not true The degree of freedom is 4 We fail to reject The p-value is 0.4646 The test Statistics is x = 0.5345

  • Score. b. Question 6 B0/3 pts 399 Det Test the claim that the mean GPA of...

    Score. b. Question 6 B0/3 pts 399 Det Test the claim that the mean GPA of night students is larger than 3.5 at the 0.025 significance level. The null and alternative hypothesis would be: H:p = 0.875 H :p > 0.875 HH <3.5 H: > 3.5 H: = 3.5 H :P < 0.875 HP 0.875 H :P < 0.875 H: > 3.5 H: < 3.5 H1: +3.5 H :p > 0.875 The test is: right-tailed two-tailed left-tailed Based on a...

  • . Question 11 3 pts Let X be normally distributed with some unknown mean and standard...

    . Question 11 3 pts Let X be normally distributed with some unknown mean and standard deviation X- o=2. The variable Z = is distributed according to the standard normal A distribution. Enter the value for A = . It is known that P(X > 14) = 0.2 What is P(Z > 7 - ) = (enter a decimal value). Use R to determine di = (round to the second decimal place).

  • Question 5 0/3 pts 5399 Details You wish to test the following claim (H) at a...

    Question 5 0/3 pts 5399 Details You wish to test the following claim (H) at a significance level of a = 0.02. Hu = 50.7 H: #50.7 You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size n = 410 with mean M = 46.7 and a standard deviation of SD = 20.7. What is the test statistic for this sample? (Report answer accurate to three decimal places.) test...

  • Question 2 2 pts An article on the effect of aerosol species on atmospheric visibility reported...

    Question 2 2 pts An article on the effect of aerosol species on atmospheric visibility reported that for 20 samples taken in the winter, the mass ratio of fine to coarse particles averaged 0.51 with a standard deviation of 0.09, and for 14 samples taken in the summer the mass ratio averaged 0.65 with a standard deviation of 0.11. Let Hi represent the mean mass ratio during the winter and let uz represent the mean mass ratio during the summer....

  • True or False: Fr questions 1-7, write either T" for True or"F" for False in the margin, next to each question number. Each question is worth 1.5 pts. let 1) The shape of the sampling...

    True or False: Fr questions 1-7, write either T" for True or"F" for False in the margin, next to each question number. Each question is worth 1.5 pts. let 1) The shape of the sampling distribution of X gets closer to the shape of the population 2) A result that is significant at the 0.01 significance level is always significant at the 0.05 3) We reject the null hypothesis whenever P-value <a distribution as n gets large. significance level. 4)...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT