A pharmaceutical company produces 500 mL bottles of a cough syrup. To ensure consistency, regulatory standards require the average fill volume to be exactly 500 mL-neither underfilled nor overfilled. Underfilling could lead to patients receiving insufficient medication, while overfilling could cause dosing inconsistencies and financial losses. To verify compliance, a quality control team randomly samples n = 45 bottles from a production batch and records their actual fill volumes. The data can be found in column 2 of the Download.csv file x=c(501,499,497,500,500,501,498,500,499,499,497,498,499,500,497,500,500,499,499,501,499,501,497,501,499,500,502,500,498,500,501,499,500,498,500,502,499,501,501,501,500,500,498,499,499) (You can copy and paste into RStudio if you like) (a) Does this data suggest that the mean fill volume differs from 500 mL? Select the appropriate statistical hypotheses. Α. Η : μ Σ 500 : B. Hoμ 500 C. Ho X 500 D. Họ : a < 500 E. Ho 500 F. HoX<500 G. Ho : |=500 ΗΑ: μ > 500 Hд: μ< 500 HA: X < 500 HA:|>500 HAμ 500 HA: X≥ 500 HA: | >500 (b) Compute the value of the test statistic. Test Statistic = (use three decimals in your answer) (c) Using the technology available to you compute the P-value. P-value = (use five decimals in your answer) (d) What is the R code for the p-value above A. pt(-abs(tcalc), df=45) B. 2*(1-pt(-abs(tcalc), df=45)) C. pnorm(-abs(tcalc), df=45) D. 1-pt(-abs(tcalc), df=45) E. 2*pt(-abs(tcalc), df=45-1) F. 2*pnorm(-abs(tcalc), df=45-1) G. pnorm(-abs(tcalc), df=45-1) (e) Testing at a = 0.05 what decision can you make about the null hypothesis? ? Ho. (e) Complete the statement, providing your conclusion below. ? one can infer that the mean fill volume is ? ✓ mL. (use at least one decimal in your answer).
A pharmaceutical company produces 500 mL bottles of a cough syrup. To ensure consistency, regulatory standards require the average fill volume to be exactly 500 mL-neither underfilled nor overfilled. Underfilling could lead to patients receiving insufficient medication, while overfilling could cause dosing inconsistencies and financial losses. To verify compliance, a quality control team randomly samples n = 45 bottles from a production batch and records their actual fill volumes. The data can be found in column 2 of the Download.csv file x=c(501,499,497,500,500,501,498,500,499,499,497,498,499,500,497,500,500,499,499,501,499,501,497,501,499,500,502,500,498,500,501,499,500,498,500,502,499,501,501,501,500,500,498,499,499) (You can copy and paste into RStudio if you like) (a) Does this data suggest that the mean fill volume differs from 500 mL? Select the appropriate statistical hypotheses. Α. Η : μ Σ 500 : B. Hoμ 500 C. Ho X 500 D. Họ : a < 500 E. Ho 500 F. HoX<500 G. Ho : |=500 ΗΑ: μ > 500 Hд: μ< 500 HA: X < 500 HA:|>500 HAμ 500 HA: X≥ 500 HA: | >500 (b) Compute the value of the test statistic. Test Statistic = (use three decimals in your answer) (c) Using the technology available to you compute the P-value. P-value = (use five decimals in your answer) (d) What is the R code for the p-value above A. pt(-abs(tcalc), df=45) B. 2*(1-pt(-abs(tcalc), df=45)) C. pnorm(-abs(tcalc), df=45) D. 1-pt(-abs(tcalc), df=45) E. 2*pt(-abs(tcalc), df=45-1) F. 2*pnorm(-abs(tcalc), df=45-1) G. pnorm(-abs(tcalc), df=45-1) (e) Testing at a = 0.05 what decision can you make about the null hypothesis? ? Ho. (e) Complete the statement, providing your conclusion below. ? one can infer that the mean fill volume is ? ✓ mL. (use at least one decimal in your answer).
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Transcribed Image Text:A pharmaceutical company produces 500 mL bottles of a cough syrup. To ensure consistency, regulatory standards require the average fill volume to be exactly 500 mL-neither underfilled nor
overfilled. Underfilling could lead to patients receiving insufficient medication, while overfilling could cause dosing inconsistencies and financial losses. To verify compliance, a quality control team
randomly samples n = 45 bottles from a production batch and records their actual fill volumes.
The data can be found in column 2 of the Download.csv file
x=c(501,499,497,500,500,501,498,500,499,499,497,498,499,500,497,500,500,499,499,501,499,501,497,501,499,500,502,500,498,500,501,499,500,498,500,502,499,501,501,501,500,500,498,499,499)
(You can copy and paste into RStudio if you like)
(a) Does this data suggest that the mean fill volume differs from 500 mL? Select the appropriate statistical hypotheses.
Α. Η : μ Σ 500
:
B. Hoμ 500
C. Ho X 500
D. Họ : a < 500
E. Ho
500
F. HoX<500
G. Ho : |=500
ΗΑ: μ > 500
Hд: μ< 500
HA: X < 500
HA:|>500
HAμ 500
HA: X≥ 500
HA: | >500
(b) Compute the value of the test statistic.
Test Statistic =
(use three decimals in your answer)
(c) Using the technology available to you compute the P-value.
P-value = (use five decimals in your answer)
(d) What is the R code for the p-value above
A. pt(-abs(tcalc), df=45)
B. 2*(1-pt(-abs(tcalc), df=45))
C. pnorm(-abs(tcalc), df=45)
D. 1-pt(-abs(tcalc), df=45)
E. 2*pt(-abs(tcalc), df=45-1)
F. 2*pnorm(-abs(tcalc), df=45-1)
G. pnorm(-abs(tcalc), df=45-1)
(e) Testing at a = 0.05 what decision can you make about the null hypothesis?
?
Ho.
(e) Complete the statement, providing your conclusion below.
?
one can infer that the mean fill volume is ?
✓
mL. (use at least one decimal in your answer).
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