Question 1068665: One of your friends knows that you are taking Advanced Placement Statistics and asks for your help wither Chemistry lab report. She has come up with five measurements of the melting point of a compound: 122.4, 121.8, 122.0, 123.0, and 122.3 degrees Celsius.
a. The lab manual asks for a 95% confidence interval estimate for the melting point. Show her how to find this estimate.
b. Explain to your friend in simple language what 95% confidence interval means.
c. Would a 90% confidence interval be narrower or wider than the 95% interval. Explain your answer to your friend.
d. The lab manual asks whether the data show sufficient evidence to call into question the established melting point of 122 degrees Celsius. State the null and alternative hypotheses and find the P-value.
e. Explain the meaning of the specific P-value to your friend.
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! 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 < 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'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 < 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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