The manufacturer of an airport baggage scanning machine claims it can handle an
ID: 3437856 • Letter: T
Question
The manufacturer of an airport baggage scanning machine claims it can handle an average of 531 bags per hour.
a.
At = 0.05 in a left-tailed test, would a sample of 16 randomly chosen hours with a mean of 511 and a standard deviation of 59 indicate that the manufacturer's claim is overstated? Choose the appropriate hypothesis.
H0: < 531. Reject H0 if tcalc > -1.753
H1: 531. Reject H1 if tcalc < -1.753
H0: 531. Reject H0 if tcalc < -1.753
H1: < 531. Reject H1 if tcalc > -1.753
b. State the conclusion.
The manufacturer of an airport baggage scanning machine claims it can handle an average of 531 bags per hour.
a.
At = 0.05 in a left-tailed test, would a sample of 16 randomly chosen hours with a mean of 511 and a standard deviation of 59 indicate that the manufacturer's claim is overstated? Choose the appropriate hypothesis.
H0: < 531. Reject H0 if tcalc > -1.753
H1: 531. Reject H1 if tcalc < -1.753
H0: 531. Reject H0 if tcalc < -1.753
H1: < 531. Reject H1 if tcalc > -1.753
b. State the conclusion.
tcalc = -1.356. There is significant evidence to reject the manufacturer’s claim. tcalc = -1.356. There is not enough evidence to reject the manufacturer’s claim.Explanation / Answer
The manufacturer of an airport baggage scanning machine claims it can handle an average of 531 bags per hour.
a.
At = 0.05 in a left-tailed test, would a sample of 16 randomly chosen hours with a mean of 511 and a standard deviation of 59 indicate that the manufacturer's claim is overstated? Choose the appropriate hypothesis.
H0: < 531. Reject H0 if tcalc > -1.753
H1: 531. Reject H1 if tcalc < -1.753
H0: 531. Reject H0 if tcalc < -1.753
H1: < 531. Reject H1 if tcalc > -1.753
b. State the conclusion.
tcalc = -1.356. There is significant evidence to reject the manufacturer’s claim.
tcalc = -1.356. There is not enough evidence to reject the manufacturer’s claim.
(Calculated t is greater than table value, do not reject for left tailed test)
t Test for Hypothesis of the Mean
Data
Null Hypothesis m=
531
Level of Significance
0.05
Sample Size
16
Sample Mean
511
Sample Standard Deviation
59
Intermediate Calculations
Standard Error of the Mean
14.7500
Degrees of Freedom
15
t Test Statistic
-1.3559
Lower-Tail Test
Lower Critical Value
-1.7531
p-Value
0.0976
Do not reject the null hypothesis
At = 0.05 in a left-tailed test, would a sample of 16 randomly chosen hours with a mean of 511 and a standard deviation of 59 indicate that the manufacturer's claim is overstated? Choose the appropriate hypothesis.
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