A quality control specialist for a restaurant chain takes a random sample of size 14 to check the amount of soda served in the 16 oz. serving size. The sample mean is 13.60 with a sample standard deviation of 1.51. Assume the underlying population is normally distributed. Find the 95% confidence interval for the true population mean for the amount of soda served.

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Answer:

The 95% confidence interval = (12.79, 14.41)

Step-by-step explanation:

The confidence interval expresses that the true mean lies within the calculated range with a particular level of confidence.

To obtain the estimate of the 95% confidence interval of the population, we'll need the standard error

Upper limit of the confidence interval = (sample mean) + (standard error)

Lower limit of the confidence interval = (Sample mean) - (standard error)

Standard error of the sample mean = (critical value) × (standard deviation of the sample mean)

(Standard deviation of the sample mean) = (standard deviation)/√n

where n = sample size = 14

(Standard deviation of the sample mean) = (1.55/√n) = (1.55/√14) = 0.4143

Critical value for 95% confidence interval = z = 1.96 (z-score from the tables)

Standard error of the sample mean = 1.96 × 0.4143 = 0.81

Upper limit of the confidence interval = (sample mean) + (standard error) = 13.60 + 0.81 = 14.41

Lower limit of the confidence interval = (Sample mean) - (standard error) = 13.60 - 0.81 = 12.79

The 95% confidence interval = (12.79, 14.41)

Hope this Helps!!!

Answer:

(12.728,14.472)

Step-by-step explanation:

As the sample size is small and population standard deviation is unknown so we use t-distribution for computing confidence interval for true population mean. Also, assumption for normality is satisfied for using t-distribution.

The 95% confidence interval for true population mean can be computed as:

[tex]xbar-t_{\frac{\alpha }{2} (n-1)}(\frac{s}{\sqrt{n} } )<mew<xbar+t_{\frac{\alpha }{2} (n-1)}(\frac{s}{\sqrt{n} } )[/tex]

where mew=μ represents true population mean.

We are given that

xbar=13.6, s=1.51 and n=14.

[tex]t_{\frac{\alpha }{2} (n-1)}=t_{\frac{\0.05}{2} (14-1)}=t_{0.025 (13)}=2.16[/tex]

[tex]13.6-2.16(\frac{1.51}{\sqrt{14} } )<mew<13.6+2.16(\frac{1.51}{\sqrt{14} } )[/tex]

[tex]13.6-2.16(0.4036 )<mew<13.6+2.16(0.4036 )[/tex]

[tex]13.60-0.8717<mew<13.60+0.8717[/tex]

[tex]12.728<mew<14.472[/tex]   (rounded to three decimal places).

Thus, the 95% confidence interval for true population mean for the amount of soda served is (12.728,14.472).