Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 months ago

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• 2 months ago

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