Do not reject H 0 . Reject H 0 . Yes. The number of thunder days appears to be l

Question

Do not reject H0. Reject H0. Yes. The number of thunder days appears to be less than 68.   There is not enough information to draw a conclusion. No. There is insufficient evidence to conclude that the number of thunder days is less than 68. Yes. The number of thunder days appears to be less than 68.   There is not enough information to draw a conclusion. No. There is insufficient evidence to conclude that the number of thunder days is less than 68. A sample of 132 tobacco smokers who recently completed a new smoking-cessation program were asked to rate the effectiveness of the program on a scale of 1 to 10, with 10 corresponding to "completely effective" and 1 corresponding to "completely ineffective". The average rating was 5.7 and the standard deviation was 4.7.

Construct a 95% confidence interval for the mean score. (5.3, 6.1) (0, 5.7) (4.9, 6.5) (5.4, 6.0) A random sample of size 16 from a normal distribution has standard deviation . Test . Use the level of significance. Do not reject H0. Reject H0. Historically, a certain region has experienced 68 thunder days annually. (A "thunder day" is day on which at least one instance of thunder is audible to a normal human ear). Over the past seven years, the mean number of thunder days is 58 with a standard deviation of 26. Can you conclude that the mean number of thunder days is less than 68? Use the level of significance. Yes. The number of thunder days appears to be less than 68.   There is not enough information to draw a conclusion. No. There is insufficient evidence to conclude that the number of thunder days is less than 68.

Explanation / Answer

2)

std error of mean =std deviation/(n)1/2 =26/(7)1/2 =9.827

therefore test stat t=(X-mean)/std error =(58-68)/9.827=-1.1076

p value =0.1740

as p value is higher then significance level

No. There is insufficient evidence to conclude that the number of thunder days is less than 68.

3)

here std error of mean =std deviation/(n)1/2 =0.409

for 95% CI ; z=1.96

hence 95% confidence interval =sample mean -/+ z*std error =4.9 ; 6..5

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