jharb65
15.07.2019 •
Mathematics
Suppose that a restaurant chain claims that its bottles of ketchup contain 24 ounces of ketchup on average, with a standard deviation of 0.8 ounces. if you took a sample of the 49 bottles of ketchup what would be the approximate 95% confidence interval for a mean number of ounces of ketchup per bottle in the sample?
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Ответ:
24+- 0.229
Step-by-step explanation:
Just took the test
Ответ:
Confidence Interval = (23.776, 24.224)
Step-by-step explanation:
Restaurant chain claims that its bottles of ketchup contain 24 ounces of ketchup on average.
⇒ Mean = 24 ounces.
Standard Deviation = 0.8
Number of bottles used for sample = 49
⇒ n = 49
Confidence level = 95%
Corresponding z value with 95% confidence level = 1.96
Now, confidence interval is given by the following expression :
Hence, Confidence Interval = (23.776, 24.224)
Ответ: