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Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 months ago

Help Please?

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  • 2 months ago
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    X: price of ticket

    f: number of tickets

    Ans(a)

    \mu= average of price ticket to Langkawi

    \bar{x}= sample average price of ticket

    H_{0}:\mu=93.00

    H: > 93.00

    n =5 ( from 20 july to 24 july)

    data follow from normal distribution. we use Z test statistic for testing of hypothesis.

    Price(RM)                   Mid value     f              x*f             x^2*f

    66.00-75.00                   70.5        15          1057.5         74553.75

    76.00-85.00                   80.5       40           3220            259210

    86.00-95.00                  90.5        60           5430            491415

    96.00-105.00              100.5        45          4522.5          454511.25

    106 above                  110.5        40          4420              488410

     

                                            total=200   total=18650  total=1768100

    mean=bar{x}=\sum x*f/\sum f=18650/200=93.25

    mean=\bar{x}=93.25

    the standard deviation is calculated as

    \sigma=\sqrt ( \sum x^2*f/\sum f-(\bar{x})^2  )}

    \sigma=\sqrt{( 1768100/200-(93.25)^2 )}

    \sigma=\sqrt{( 8840.5-8695.5625 )}

    \sigma=\sqrt{( 144.9375 )}

    \sigma=12.0389

    the Z test statistics is given by

    Z=\sqrt{n}*(\bar{x}-\mu)/\sigma

    Z=\sqrt{5}*(93.25-93.00)/12.0389

    Z=2.236*(0.25)/12.0389

    Z=0.0464

    the tabulated value of Z at 5% level of significance = 1.645

    Conclusion:

    since calculated Z is much less than tabulated Z . we do not reject the null hypothesis and conclude that sales manager cannot say that the average price of ticket to LangKawi during this particular period is above the average of price 93.00

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