<PRE><B>Question 1068665</B>
We are to use the sample mean to estimate the population mean for the melting point of the compound
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sample mean = (122.4 + 121.8 + 122.0 + 123.0 + 122.3) / 5 = 122.3
sample standard deviation = square root( (1/5) * summation from i = 1 to 5 of (x(i) - 122.3)^2 ) = 0.45826
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standard error(SE) = sample standard deviation / square root(sample size) = 0.45826 / square root(5) = 0.2049
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margin of error(ME) = critical value(CV) * SE
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since the sample size is &lt; 5, we use the student t-value as our CV
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degrees of freedom(DF) = 5 - 1 = 4
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we are asked to calculate a 95% confidence interval for the population mean
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alpha(a) = 1 - (95/100) = 0.05
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critical probability(p*) = 1 - (a/2) = = 1 - 0.025 =  0.975
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consulting t distribution table for DF = 4 and p* = 0.975, we have CV = 2.776
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ME = 2.776 * 0.2049 = 0.5688
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95% confidence interval(CI) is sample mean + or - ME
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a) 95% CI = 122.3 + or - 0.5688, (121.7312, 122.8688)
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b) The 95% CI tells us that the population mean for 
the compound&#39;s melting point is greater than 121.7312 and
less than 122.8688
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c) The 90% CI will be narrower than the 95% CI, note that
the larger the probability the wider the CI becomes
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I will calculate the 90% CI so you can see the relationship
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a = 1 - (90/100) = 0.10
p* = 1 - (0.10/2) = 0.95
DF = 4
CV = 2.132
ME = 2.132 * 0.2049 = 0.4368
90% CI = 122.3 + or - 0.4368, (121.8632, 122.7368)
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d) H(0): x not = 122, H(1): x = 122, note H(0) is null hypothesis
In this case we use a two-tailed test because H(1) includes an = sign
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we are considering the 95% CI, so the significance level is 0.05
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The p-value for our CV = 2.776 is 2 * (0.025) = 0.05
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our p-value 0.05 = significance level of 0.05
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We accept our H0, since the p-value is NOT &lt; 0.05
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e) Our calculated p-value is the probability of a sample of 5 having a sample mean of at least 122.3 Celsius given that the population mean is not = 122 Celsius
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