Mathematics, 30.11.2019 01:31 fish64

The manufacturer of an airport baggage scanning machine claims it can handle an average of 530 bags per hour. (a-1) at α = .05 in a left-tailed test, would a sample of 16 randomly chosen hours with a mean of 510 and a standard deviation of 50 indicate that the manufacturer’s claim is overstated? choose the appropriate hypothesis.

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Answer from: Jefferson09

Null hypothesis: $\mu \geq 250$

Alternative hypothesis: $\mu < 250$

$p_v =P(t_{15}$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significantly lower than 530 at 5% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X=510$ represent the mean for the sample

s=50 represent the standard deviation for the sample

n=16 sample size

$\mu_o =530$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

p_v represent the p value for the test (variable of interest)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the claim thet the handle on average is 530 bags per hour:

Null hypothesis: $\mu \geq 530$

Alternative hypothesis: $\mu < 530$

We don't know the population deviation and the sample size is lees than 30, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

3) Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{510-530}{\frac{50}{\sqrt{16}}}=-1.60$