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Data analysis in business Tutorial 4

Question A E-grocers are companies that sell groceries online. A potential e-grocer analyzed the market and determined that to be profitable the average order needs to exceed $80. To determine whether egrocery would be profitable in a new city, the company offered trial services in the city and recorded the size of the orders (in $) for a random sample of 85 customers. Use the data in T4.xlsx to help the company to decide whether e-grocery would be profitable in this city. 1. Form the null and the alternative hypotheses. Let µ denote the mean order size, the null and the alternative are H0 : µ = 80 HA : µ > 80 2. Create a histogram and comment on the shape of the histogram. Click Insert - Chart -Histogram to create the histogram. The histogram is approximately bell shaped. 3. Obtain the sample mean, standard error of sample mean, and the sample standard deviation from using Excel. Using Data analysis - Descriptive Statistics we obtain ¯ = 82.44, s = 11.18, SE(X) ¯ = 1.21 X 4. Assume that the order sizes are normally distributed, can we infer at the 1% significance level that an e-grocery will be profitable in this city? Use the critical value approach. Test statistic and null distribution: T =

X¯ − 80 ¯ ∼ t84 SE( X)

Value of the test statistic from sample: t=

¯ − 80 82.44 − 80 X = 2.01 ¯ = SE(X) 1.21

Critical value for the test: t0.99 = 2.37 Since the t statistic is smaller the the critical, we do not reject the null at the 1% level. 1

Question B A recent study showed that 79% of companies allow employee to work remotely. Suppose a researcher believes that in accounting companies this figure is lower. The researcher randomly selects 415 accounting companies, and through interviews determines that 303 of these companies allows remote working. Using a 1% level of significance, test if there is enough evidence to conclude that the proportion of accounting companies that allows remote working is significantly lower than 79%. Step 1: set up the hypotheses: H0 : π = 0.79 HA : π < 0.79 Step 2: α = 0.01 Step 3: Use sample proportion p to estimate π, and use the test statistic and null distribution given by Z= p

p − π0 π0 (1 − π0 )/n

∼ N (0, 1)

Step 4: the sample estimate p = 303/415 = 0.7301; the value of the test statistic from sample is 0.7301 − 0.79 z= p = −2.995 0.79 × 0.21/415 Step 5: the critical value is z0.01 = −2.33. Since z < z0.01 , we reject the null. In other words, there is sufficient evidence to conclude that the proportion of accounting companies that allows remote working is significantly lower.

Question C Type C batteries have a nominal diameter of 26.2mm. To ensure they fit properly into all toys and appliances that require C batteries, the industry standard allows a standard deviation of 0.5mm in the production process. After a repair, a machine produced a batch of 12 batteries. The sample mean of this batch is 26.2mm and the sample standard deviation is 0.6mm. The managers in the factory worry the variance may be too high and they are considering sending the machine for further repairs. 1. Form a null and an alternative hypotheses to help the managers decide whether there’s enough statistical evidence to suggest that the machine needs further repairs. The null and alternatives are H0 : σ = 0.5 HA : σ > 0.5

2

or H0 : σ 2 = 0.25 HA : σ 2 > 0.25 2. What are the type 1 and type 2 errors in this case? Type 1 error in this case is sending the machine for repairs when the machine is fine. Type 2 error is letting the machine keep operating when the machine does produce defective products more than usual. 3. Suppose that manager A is more worried about the cost from further repairing the machine. Manage B is more worried about the cost from having more than usual defective products. Which manager would use a lower significance level for the hypotheses test? Manager A prefers a smaller type 1 error than manage B; and manager B prefers a smaller type 2 error than manager A. Since significance level is tie to the probability of type 1 error, manager A (the one who has lower tolerance for type 1 error) would use a lower significance level for the test. 4. Set the significance level to 5%. Use the p-value approach to decide whether or not to reject the null? The null distribution of test statistic: Q=

11 × s2 2 ∼ χ11 0.25

The value of test statistic from the data: q=

11 × 0.36 = 15.84 0.25

The p-value = Pr(Q > q) = 1 − Pr(Q < q) =1-CHISQ.DIST(15.84,11,1)≈ 0.147 Since the p-value is larger than 5%, we do not reject the null. In other words, there is insufficient evidence to conclude that the machine needs further repairs.

Question D Suppose that in past years the average weekly rent in Caulfield has been $322.40 and the prices follow a normal distribution. A real estate investor wants to determine whether the figure has changed now. The investor hires a researcher who randomly samples 19 apartments in Caulfield and finds that the mean rent is $316.70, with a standard deviation of $12.90. 1. Compute and interpret the 95% confidence interval for the average weekly rent in Caulfield. For the 95% confidence interval, we use the 2.5th and 97.5th percentiles: t0.025,18 = −2.101 and t0.975,18 = 2.101 3

¯ = SE(X)

12 √ .9 19

≈ 2.96

The confidence interval is (316.7 − 2.101 ∗ 2.96, 316.7 + 2.101 ∗ 2.96) = (310.5, 322.9) We are 95% confident that the average weekly rent in Caulfield is between 310.5 and 322.9. 2. From the 95% confidence interval, do you reject the null that rent has not changed at the 5% level? Since the value under the null (322.4) is within the 95% confidence interval, we do no reject the null at the 5% level. In other words, there is insufficient evidence to conclude that the average weekly rent has changed. 3. Compute and interpret the 95% prediction interval for the weekly rent in Caulfield. Let X denote the weekly rent in Caulfield, from the sample, we have E(X) = 316.7, and SD(X) = 12.9 For the 95% prediction interval, we also use t0.025,18 = −2.101 and t0.975,18 = 2.101 The 95% prediction interval is (316.7 − 2.101 ∗ 12.9, 316.7 + 2.101 ∗ 12.9) = (289.6, 343.8). There is a 95% probability that the weekly rent in a randomly chosen apartment in Caufield is between 289.6 and 343.8.

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