Ask Your Teacher A journal article reports that a sample of size 5 was used as a basis for calculating a 95% CI for the true average natural frequency (Hz) of delaminated beams of a certain type. The resulting interval was (229.266, 233.002). You decide that a confidence level of 99% is more appropriate than the 95% level used. What are the limits of the 99% interval? [Hint: Use the center of the interval and its width to determine x and s.] (Round your answers to three decimal places.)

The 99% confidence interval is (228.035, 234.233).

Step-by-step explanation:

The sample size selected to compute the 95% confidence interval for the true average natural frequency (Hz) of delaminated beams of a certain type is

n = 5

The sample size is very small and the population standard deviation is also not known.

So, we will use t-interval for the confidence interval.

The (1 - α)% confidence interval for the true mean is:

The 95% confidence interval for the true average natural frequency (Hz) of delaminated beams of a certain type is:

(Upper limit, Lower limit) = (229.266, 233.002).

Compute the value of sample mean as follows:

The critical value of t for α = 0.05 and n = 5 is:

*Use a t-table.

Compute the value of sample standard deviation as follows:

The critical value of t for α = 0.01 and n = 5 is:

*Use a t-table.

Construct the 99% confidence interval as follows:

Thus, the 99% confidence interval is (228.035, 234.233).

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐