Compare Mean with Expected Value You purchase a chemical substance which is cert

Question

Compare Mean with Expected Value

You purchase a chemical substance which is certified to be 98.3% pure. Your own repeated analysis for the percent-purity of this substance produces 98.1, 96.4, 94.3, 97.3, 97.7 and 95.2. Give the 95% confidence interval by providing the mean and the corresponding uncertainty. mean measured % purity = ± (for n=6 measurements and 95% confidence)

Do your results differ from the expected result at the 95% confidence level?

Yes or No

You repeat the analysis just once more and get 99.2%. Again give the 95% confidence interval by providing the new mean and the corresponding uncertainty. mean measured % purity = ± (for n=7 measurements and 95% confidence)

Would your conclusion change? No Yes

Explanation / Answer

Average = 96.5

Calculation

M = 96.5
t = 1.96
sM = (1.4752/6) = 0.6

= M ± Z(sM)
= 96.5 ± 1.96*0.6
= 96.5 ± 1.18

Ye the results differ from the expected result of 98.3%

Now we add 99.2

Mean = 96.88

Sd = 1.7

Calculation

M = 96.88
t = 1.96
sM = (1.72/7) = 0.64

= M ± Z(sM)
= 96.88 ± 1.96*0.64
= 96.88 ± 1.2594

the conclusion does not change

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