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12 August, 00:23

A sample of 270 students who were taking online courses were asked to describe their overall impression of online learning on a scale of 1-7, with 7 representing the most favorable impression. The average score was 5.81, and the standard deviation was 0.99. Construct an 80% confidence interval for the mean score. Round the answers to two decimal places l Ask An 80% confidence interval for the mean score is [ ] <.

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  1. 12 August, 00:49
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    An 80% confidence intervals are (5.73,5.88)

    Step-by-step explanation:

    Given sample size (n) is = 270

    The average score (μ) = 5.81

    and standard deviation σ = 0.99.

    80% confidence interval:-

    The 80% confidence interval of the z - value is 1.28 (from z-table)

    An 80% confidence interval is defined by sample mean ± 1.28 standard error

    that is μ ± 1.28 σ/√n

    now substitute values (5.81 ± 1.28 (0.99/√270))

    (5.81 - 1.28 (0.0602),5.81 + 1.28 (0.0602)

    (5.81 - 0.0770,5.81 - 0.0770)

    (5.73,5.88)

    Conclusion:-

    An 80% confidence intervals are (5.73,5.88)

    Therefore the population mean 5.81 lies between (5.73,5.88) at 80% confidence intervals
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