Ellie, an office manager, needs to find a courier to deliver a package. The first courier she is considering charges a fee of $1
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Answer:
H1 : μ < 553. Reject H1 if tcalc > –1.753.
There is not enough evidence to reject the manufacturer’s claim.
Step-by-step explanation:
We are given that the manufacturer of an airport baggage scanning machine claims it can handle an average of 553 bags per hour.
A sample of 16 randomly chosen hours with a mean of 533 and a standard deviation of 47 is given.
Let = <u><em>average bags that an airport baggage scanning machine can handle.</em></u>
SO, Null Hypothesis, :
553 bags {means that the manufacturer’s claim is not overstated}
Alternate Hypothesis, :
< 553 bags {means that the manufacturer’s claim is overstated}
The test statistics that would be used here <u>One-sample t-test statistics</u> as we don't know about population standard deviation;
T.S. = ~
where, = sample mean = 533 bags
s = sample standard deviation = 47
n = sample of hours = 16
So, <u><em>the test statistics</em></u> = ~
= -1.702
The value of t test statistics is -1.702.
<u>Now, at 0.05 significance level the t table gives critical value of -1.753 at 15 degree of freedom for left-tailed test.</u>
Since our test statistic is more than the critical value of t as -1.702 > -1.753, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u><em>we fail to reject our null hypothesis</em></u>.
Therefore, we conclude that the manufacturer’s claim is not overstated and an airport baggage scanning machine can handle an average of 553 bags per hour.