Please Solve the Problems. Construct a confidence interval for µd, the mean of t

Question

Please Solve the Problems.

Construct a confidence interval for µd, the mean of the differences d for the population of paired data. Assume that the population of paired differences is normally distributed

Given the linear correlation coefficient r and the sample size n, determine the critical values of r and use your finding to state whether or not the given r represents a significant linear correlation. Use a significance level of 0.05.

Using the sample paired data below, construct a 90% confidence interval for the population mean of all differences x y x 4.0 6.9 4.7 4.1 7.2 y 3.7 5.7 4.3 4.8 4.9 B) -0.31 ud 1.71 0.07 ud 1.47 0.22 Hd 7.48 DO -0.37 ud 1.77

Explanation / Answer

Construct a confidence interval for µd, the mean of the differences d for the population of paired data. Assume that the population of paired differences is normally distributed.

We have to construct 90% confidence interval for the population mean of all differences x-y.

We can construct confidence interval for population mean using MINITAB.

steps :

ENTER data into MINITAB sheet --> STAT --> Basic statistics --> Paired t --> Samples in columns --> First column : x --> Second column : y --> Options --> Confidence level : 90% --> Test mean : 0.0 --> Alternative : not equal --> ok --> ok

Paired T-Test and CI: x, y

Paired T for x - y

N Mean StDev SE Mean
x 5 5.38000 1.55145 0.69383
y 5 4.68000 0.74297 0.33226
Difference 5 0.700000 1.120268 0.500999

90% CI for mean difference: (-0.368053, 1.768053)
T-Test of mean difference = 0 (vs not = 0): T-Value = 1.40 P-Value = 0.235

90% confidence interval for population mean is (-0.37, 1.77).

Here option D) is correct.

Given the linear correlation coefficient r and the sample size n, determine the critical values of r and use your finding to state whether or not the given r represents a significant linear correlation. Use a significance level of 0.05.

Here we have to test the hypothesis that,

H0 : Rho = 0 Vs H1 : Rho not= 0

where Rho is population correlation.

Assume alpha = 0.05

r = 0.45

n = 25

Now we have to find critical value for t-distribution.

Critical value we can find using critical value of Pearson correlation.

Critical value for the test are + or - 0.396.

Here r > critical value

Reject H0 at 5% level of significance.

Conclusion : There is significant linear correlation.

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