A random sample of 10 digital photos are taken outside in July have a mean size

Question

A random sample of 10 digital photos are taken outside in July have a mean size of 2.2 MB with a standard deviation of 0.3 MB. A random sample of 8 digital photos taken inside in January with the same camera have a mean size of 1.8 MB and a standard deviation of 0.4 MB.

a) Construct a 95% confidence interval for the mean size of July photos taken outside

b) What assumption(s) are required for this interval to be valid?

c) Would a 90% interval be narrower or wider?

d) Is the mean size of photos taken outside in July different from that of photos taken inside in January? i. state the hypotheses ii. calculate the test statistic iii. find the P value, using bound if necessary. state the table used and show values from the table. iv. give a conclusion in the context of the problem. v. state what extra assumptions beyond those in part b) are required for the test to be valid.

Explanation / Answer

a.) µ = 2.2; s = 0.3; n = 10

SE = s / n = 0.3/10 = 0.0948

t = 1.833

ME = 1.833*0.0948 = 0.1737

CI = [2.2 ± 0.1737] = [2.026, 2.3737]

b) Each sample is an independent random sample.

The distribution of the response variable follows a normal distribution.

The population variances are equal across responses for the group levels.

c) A 90% Confidence interval will be narrower than a 95% confidence interval

d) i.) H0: 1 = 2
Ha: 1 2

ii.) SE = sqrt[ (s12/n1) + (s22/n2) ] = 0.1702

  DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] } ~ 14

  t = [ (x1 - x2)] / SE = 2.34

iii.) This is a two tailed test. Hence, p-value = 0.0346

iv.) Taking significance level as 95%. The result is significant at p < 0.05

Since, 0.0346 < 0.05. Hence, we cannot reject the null hypothesis.

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