On day two of a study on body temperatures, 106 temperatures were taken. Suppose that we only have the first 10 temperatures to work with. T

Question

On day two of a study on body temperatures, 106 temperatures were taken. Suppose that we only have the first 10 temperatures to work with. The mean and standard deviation of these 10 scores were 98.44o F and 0.30o F, respectively. Construct a 95% confidence interval for the mean of all body temperatures.

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Kiệt Gia 1 year 2021-08-13T23:23:06+00:00 1 Answers 40 views 0

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    2021-08-13T23:24:22+00:00

    Answer:

    The 95% confidence interval = (98.225, 98.655)

    Step-by-step explanation:

    The formula for Confidence Interval =

    Mean ± z × standard deviation/√n

    Where Mean = 98.44°F

    Standard deviation = 0.30°F

    n = number of samples is 10

    Because our number if samples is small, we use the t score confidence interval formula

    Mean ± t × standard deviation/√n

    Degrees of Freedom = n – 1

    = 10 – 1 = 9

    We find the t score of 95% confidence interval and degrees of freedom 9

    = 2.262

    Hence,

    Confidence interval = 98.44 ± 2.262 × 0.30/√10

    = 98.44 ± 0.214592162

    Confidence Interval

    98.44 – 0.214592162

    = 98.225407838

    ≈ 98.225

    98.44 + 0.214592162

    = 98.654592162

    ≈ 98.655

    The 95% confidence interval = (98.225, 98.655)

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