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Maru [420]
3 years ago
15

The sound intensity of a certain type of food processor in normally distributed with standard deviation of 2.9 decibels. If the

measurements of the sound intensity of a random sample of 9 such food processors showed a sample mean of 50.3 decibels, find a 95% confidence interval estimate of the (true, unknown) mean sound intensity of all food processors of this type
Physics
1 answer:
Maru [420]3 years ago
6 0

Answer: (48.41,\ 52.19)

Explanation:

The confidence interval for population mean is given by :-

\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}

Given : Sample size : n=9

Sample mean : \ovreline{x}=50.3\text{ decibels}

Standard deviation : \sigma=2.9\text{ decibels }

Significance level : \alpha=1-0.95=0.05

Critical value : z_{\alpha/2}=z_{0.025}=1.96

Now, the 95% confidence interval estimate of the (true, unknown) mean sound intensity of all food processors of this type :-

50.3\pm (1.96)\dfrac{2.9}{\sqrt{9}}\\\\\approx50.3\pm1.89\\\\=(50.3-1.89,\ 50.3+1.89)=(48.41,\ 52.19)

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