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Question 18
In a region of England, the government decides to use an advertising campaign to encourage people to eat more healthily. Before the campaign, the mean consumption o... show full transcript
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Answer
Let us formulate our null and alternative hypotheses:
Next, we calculate the test statistic using the formula: z = rac{ar{x} - ext{μ}}{rac{ ext{σ}}{ ext{√n}}} Where:
Thus, we find: z = rac{65.4 - 66.5}{rac{21.2}{ ext{√750}}} Calculating the standard error: ext{SE} = rac{21.2}{ ext{√750}} ≈ 0.732 Now, substituting back: z ≈ rac{-1.1}{0.732} ≈ -1.50
We compare the calculated z-value to the critical z-value at the 10% level of significance, which is -1.28 for a one-tailed test. Since -1.50 < -1.28, we reject H₀.
Conclusion: There is sufficient evidence at the 10% level of significance to suggest that the advertising campaign has reduced the consumption of chocolate per person per week.
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