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Data analysis: hypothesis testing
Data analysis: hypothesis testing

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5.4 Finding Finding x-bar

Building upon our previous knowledge of z-tests, we will now apply this statistical method to answer a new question in the field of marketing.

Consider the following scenario:

Historical records show that customers need an average of 10 seconds of exposure to television advertising commercials before being influenced, with a standard deviation of 1.6 seconds. A marketing manager believes it now takes longer to influence customer behaviour. To support this claim, he plans to sample 100 customers. We need to determine how long customers’ exposure needs to be to influence behaviour in order to justify the marketing manager’s claim with 90% confidence.

This question is vital because marketing managers are not particularly concerned with abstract statistical concepts like z critical value, rejection region or fail to reject region. Instead, they need practical, actionable insights to guide their advertising strategies.

Define the Hypotheses

  • H0: μ ≤ 10 seconds (The average time to influence customers has not increased)
  • H1: μ > 10 seconds (The average time to influence customers has increased)

Set the Confidence Level and Critical Value

  • Confidence Level: 90% (α = 0.10 for a one-tailed test)
  • Z critical value: using excel function =NORM.S.INV(0.90) = 1.28

Calculate the Sample Mean

cap z equals x bar minus mu divided by left parenthesis sigma divided by square root of n right parenthesis

x macron  = sample mean

mu  = population mean

sigma  = population standard deviation

n = sample size

We can input these values into this formula to solve the value of x.

Step 1:

1.28 equals x bar minus 10 divided by left parenthesis 1.6 divided by square root of 100 right parenthesis

Step 2:

1.28 times 1.6 divided by square root of 100 equals x bar minus 10

Step 3:

left parenthesis 1.28 times 1.6 divided by square root of 100 right parenthesis plus 10 equals x bar

Step 4:

x bar equals 10.2

To justify the marketing manager’s claim with 90% confidence, the sample of 100 customers

must show an average time to influence customer behaviour of more than 10.2 seconds.

In practical terms:

  • If the sample average is 10.2 seconds or less, we do not have sufficient evidence to reject the null hypothesis. This means we cannot conclude that the average time to influence customers has increased.
  • If the sample average exceeds 10.20 seconds, we have statistical evidence at a 90% confidence level to reject the null hypothesis and support the marketing manager’s claim that it now takes longer to influence customer behaviour.
  • The marketing team should conduct their survey of 100 customers and calculate the average time it takes to influence their behaviour. If this average is greater than 10.20 seconds, it provides support for the marketing manager’s assertion.
  • This result suggests that an increase of more than 0.20 seconds in the average time could be statistically significant given the sample size and variance in the data.
  • The marketing team should consider practical significance alongside statistical significance. While an increase to just over 10.20 seconds is statistically significant, they should evaluate whether this small increase has meaningful implications for their advertising strategies.

Activity 5: Determine the sample mean to reject H0

Timing: Allow around 40 minutes for this activity.

According to store records, customers are persuaded to purchase products with an average price discount of 50% in marketing promotions, with a standard deviation of 5.3%. The market research team believe that the company should offer more than 50% price discounts to motivate customers’ purchase intentions. To test this claim, the team plan to survey 1,000 customers. We need to determine the decision rule at a 99% confidence level to accept the market research team’s claim.

Use the textbox below to show your calculations.

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Answer

Step 1:Define the Hypotheses

  • H0: μ ≤ 50% (Customers are persuaded by price discounts of 50% or less)
  • H1: μ > 50% (Customers require price discounts greater than 50% to be persuaded)

Step 2: Identify the z-score and rejection region.

For a one-tailed test at 99% confidence level (α = 0.01), using the Excel function =NORM.S.INV(0.99), we get a z critical value of 2.326 ≈ 2.33

Step 3: Identify all the values.

mu equals 0.50 (expressed as a decimal)

sigma equals 0.053 (5.3% expressed as a decimal)

n equals 1,000

z equals 2.33 (approximate z critcail value for 99% confidence level)

Step 4: Calculate the Sample Mean

2.33 equals x bar minus 0.5 divided by left parenthesis 0.053 divided by square root of 1000 right parenthesis

Then multirelation x macron almost equals 0.504 equals 50.4 percent

Step 5: State the Decision Rule

The market research team should reject the null hypothesis (and accept their claim that customers require more than 50% discount) if the survey of 1,000 customers shows that, on average, customers are persuaded to purchase products when the price discount is greater than 50.4%.