Question

Use-the-pollooing Sample to test- whether the Pophlation Mean S ignificantly different from 5.0176 gm 4 Null, AIL α=.05 Cr ti
0 0
Add a comment Improve this question Transcribed image text
Answer #1

H0 : mu = 5.0976
Ha: mu not equasl to 5.0976

Critical value ;

-1.96 , 1.96

From the given data,

xbar = 8.20
Sigma = 2.1679

Observed difference = xbar - mu
= 8.20 - 5.0976
= 3.1024


std.error = (sigma/sqrt(n))
= ( 2.1679/sqrt(5))
= 0.9695

test statistics:

t = obs. diff/std.error
= 3.1024/0.9695

= 3.20

p value = 0.0329

Reject H0

yes outcome will change at 0.01

Add a comment
Know the answer?
Add Answer to:
Use-the-pollooing Sample to test- whether the Pophlation Mean S ignificantly different from 5.0176 gm 4 Null,...
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
  • 1. A researcher intends to use a Chi2 test for Goodness of Fit to determine whether...

    1. A researcher intends to use a Chi2 test for Goodness of Fit to determine whether there is any preference between 10 brands of cookies. What is the critical Chi2 value for a sample of n = 300 participants and p < .01? A. 20.48 B. 11.07 C. 16.92 D. 18.31 E. 21.67 2. A researcher conducted a study on a sample of n = 80 undergraduates to evaluate preferences among students for four different textbooks for the Introduction to...

  • 3 Lately, Jude has been feeling a little sad and depressed. He suspects he has been listening to ...

    3 Lately, Jude has been feeling a little sad and depressed. He suspects he has been listening to too many sad songs in his extensive iTunes collection. Jude takes a random sample from his thousands of downloaded tunes and classifies ech song as 'Sad' or Not Sad Estimate the population proportion of sad songs in Jude's collection using the following sample. Sad 160 Not Sad 40 Level of Confidence-99% Sploynit (written interpretation) Point Estimate - Critical Value Standard Errors Bound...

  • From a nationwide study, we know that the mean diastolic blood pressure is 66.1 mm gH...

    From a nationwide study, we know that the mean diastolic blood pressure is 66.1 mm gH for children aged 5-6 years of age, and that the measurements are normally distributed. Blood pressure measurements were taken on 11 children aged 5-6 years living in a specific community to determine whether their living conditions resulted in a difference in mean blood pressure. For these children the average diastolic blood pressure was found to be 60.4 mm Hg with standard deviation 8 mm...

  • Use a t-test to test the claim about the population mean μ at the given level...

    Use a t-test to test the claim about the population mean μ at the given level of significance α using the given sample statistics. Assume the population is normally distributed. Claim: μ# 25; α:0.05 Sample statistics: x 29.9, s-44, n 11 What is the value of the standardized test statistic? The standardized test statistic is(Round to two decimal places as needed.) What is the P-value of the test statistic? P-value (Round to three decimal places as needed.) Decide whether to...

  • sample mean = 213.4552 sample Standard deviation = 44.81542 N=50 alpha = .05 SEM = 6.337857477...

    sample mean = 213.4552 sample Standard deviation = 44.81542 N=50 alpha = .05 SEM = 6.337857477 For each of the following hypothesis testing problems, manually calculate the t-statistic, use the 5% level of significance (alpha = 0.05), determine the rejection region, determine the p-value of the t-test, use the 95% confidence interval in part (c) to make a decision about whether or not to reject the null hypothesis. Test the null hypothesis that the true mean is 225 versus the...

  • 2. The following table shows data on the number of parties attended by different college majors....

    2. The following table shows data on the number of parties attended by different college majors. Test whether there is a difference in the number of parties attended by conducting a Chi-Square Goodness-of-Fit test. Set alpha α = .05 Engineering English Math Psychology Sociology 2 7 4 7 5 Step 1: State the null and research hypotheses in symbols Step 2: Find the critical value (1 point) Step 3: Compute the appropriate test statistic Step 4: Make a decision about...

  • 1.Using an appropriate statistical test decide whether you can accept or reject your null hypothe...

    1.Using an appropriate statistical test decide whether you can accept or reject your null hypothesis regarding the hiv.csv data from the lab test exercise with an alpha of 0.05. Select one: a. Accept b. Reject 2.Calculate the lower value (2dp) of a 95% confidence interval from the hiv.csv data from the lab test exercise. 3.Using a Shapiro-Wilks normality test, decide whether you can accept or reject the assumption of normality of a parametric test with the hiv.csv data from the...

  • Difference of Means Test. A sample of seniors taking the SAT in Connecticut in 2015 revealed...

    Difference of Means Test. A sample of seniors taking the SAT in Connecticut in 2015 revealed the following results for the math portion of the exam by Gender. The conclusion of a hypothesis test that the mean the SAT Math scores for males and females are different at the .01 level of alpha is to fail to reject the Null Hypothesis. Note: the sample sizes are large and I will allow you to use a z-value for the critical value....

  • Suppose you want to determine whether there is a significant difference in mean test scores for...

    Suppose you want to determine whether there is a significant difference in mean test scores for females and males. The test is out of 600 points. Using the following hypotheses: Ho: U1 - U2 = 0 HA: U1 - U2 (not equal) 0 And alpha of 0.05 you obtain the following results t-test Two-sample assuming unequal variances Females   Males Mean 525 487 Variance 3530.8 2677.818182 Observations 16 12 Hypothesized Mean Difference 0 df 25 t stat 1.803753 P (T<=t) one...

  • on 6 A sample n.25 is selected from a population with mean, u = 50, 0=...

    on 6 A sample n.25 is selected from a population with mean, u = 50, 0= 10, a researcher calculated the sample mean, M - 55. Assuming the researcher was performing a two-tailed test, is the sample mean significantly different from the population mean? For the test, a -.05. ed out of Select one: O a.p=0.4938, no difference between the sample mean and the population mean O b.p = 0.0062, significant difference between the sample mean and the population mean...

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