Suppose the time that it takes a certain large bank to approve a home loan is Normally distributed with mean (in days) μ and standard deviation σ=1. The bank advertises that it approve loans in 5 days, on average, but measurements on a random sample of 500 loan applications to this bank gave a mean approval time of = 5.3 days. Is this evidence that the mean time to approval is actually longer than advertised? To answer this, test the hypotheses H0:μ=5, Ha:μ>5 at significance level α=0.01. You conclude that: a. Ha should be rejected. b. there is a 5% chance that the null hypothesis is true. c. H0 should be rejected. d. H0 should not be rejected.

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Mathematics

Moss

17 June, 06:41

Suppose the time that it takes a certain large bank to approve a home loan is Normally distributed with mean (in days) μ and standard deviation σ=1. The bank advertises that it approve loans in 5 days, on average, but measurements on a random sample of 500 loan applications to this bank gave a mean approval time of = 5.3 days. Is this evidence that the mean time to approval is actually longer than advertised? To answer this, test the hypotheses H0:μ=5, Ha:μ>5 at significance level α=0.01. You conclude that: a. Ha should be rejected. b. there is a 5% chance that the null hypothesis is true. c. H0 should be rejected. d. H0 should not be rejected.

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Julien Randall · Answer 1

(c) H0 should be rejected

Step-by-step explanation:

Null hypothesis (H0) : population mean is equal to 5

Alternate hypothesis (Ha) : population mean is greater than 5

Z = (sample mean - population mean) : (sd/√n)

sample mean = 5.3, population mean = 5, sd = 1, n = 500

Z = (5.3 - 5) : (1/√500) = 0.3 : 0.045 = 6.67

Using the normal distribution table, for a one tailed test at 0.01 significance level, the critical value is 2.326

Conclusion:

Since 6.67 is greater than 2.326, reject the null hypothesis (H0)