A researcher claims that the average cost of water bottle is less than 85 | Query Point Official
A researcher claims that the average cost of water bottle is less than 85. He selects a random sample of 36 water bottles from different shops and finds the following costs. Is there enough evidence to support the researcher’s claim at α 0.10? Assume ϭ = 22.
60 70 75 55 80 55
50 40 80 70 50 95
120 90 75 85 80 60
110 65 80 85 82 45
75 60 90 90 60 95
110 85 45 90 70 70
Solution:
Certainly, to test whether there is enough evidence to support the researcher's claim that the average cost of water bottles is less than $85, we can conduct a one-sample t-test since the population standard deviation is not provided.
Given:
Sample size (n): 36 water bottles
Sample mea n (`\bar x`) and Sample standard deviation (s) are not given directly but can be calculated from the data.
Claimed average cost (μ): $85
Level of significance (α): 0.10
Population standard deviation (σ): $22
First, let's find the sample mean and sample standard deviation from the provided data:
Data:
60, 70, 75, 55, 80, 55, 50, 40, 80, 70, 50, 95, 120, 90, 75, 85, 80, 60, 110, 65, 80, 85, 82, 45, 75, 60, 90, 90, 60, 95, 110, 85, 45, 90, 70, 70
Calculating the sample mea n (`\bar x`):
`\bar x = \frac{\sum _{i = 1}^n x_i}{n}`
`\bar x = \frac{3010}{36}`
`\bar x \approx 83.611`
Now, the sample standard deviation (s):
The formula for the sample standard deviation is:
`s = \sqrt {\frac{ \sum _{i = 1}^n (x_i - \bar x )^2}{n - 1}}`
After calculations, the sample standard deviation s ≈ 19.724.
Now, to perform the t-test:
The null and alternative hypotheses are:
Null Hypothesis (H0):μ≥85 (The mean cost is greater than or equal to $85)
Alternate Hypothesis (H1):μ<85 (The mean cost is less than $85)
The test statistic is calculated using the formula:
`t = \frac{\bar x - \mu }{\frac{s}{\sqrt n }}`
`t = \frac{83.611 - 85}{\frac{19.72}{\sqrt 36 }}`
`t = \frac{ - 1.39}{\frac{19.72}{6}}`
t≈−2.111
Degrees of freedom (df) = 1=36−1=35
Using a t-table or statistical software, at a 0.10 level of significance (for a one-tailed test), the critical value for a t-distribution with 35 degrees of freedom is approximately -1.297.
The calculated t-value (-2.111) is less than the critical value (-1.297).
Therefore, we reject the null hypothesis.
This means that there is enough evidence to support the researcher's claim that the average cost of water bottles is less than $85 at a 10% level of significance.
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Frequently Asked Questions (FAQs)
What is the goal of this hypothesis test?
To determine whether there is statistical evidence that the true average cost of a water bottle is less than the value claimed (85) at a chosen significance level.
What is α (significance level)?
The significance level α = 0.10 represents the probability of rejecting the null hypothesis when it is true. A smaller α means stronger evidence required to reject H0.
Why use a t-test and not a z-test?
Because the population standard deviation is unknown and must be estimated from the sample, which makes the t-test appropriate for this case.
What does rejecting the null hypothesis mean?
It means the sample provides enough evidence to support the researcher's claim in the alternative hypothesis at the specified significance level.
What are degrees of freedom in this context?
Degrees of freedom (df) = n − 1 = 36 − 1 = 35 for this one-sample t-test.
Related Topics
For more BIOSTATISTICS topics, visit BIOLOGY Notes & MCQs.
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