Question

After a lot of hard work, a studio releases a new video game. The studio believes that the game will be liked by players; in particular, the studio claims that the mean player rating, ( mu ), will be higher than 90 . In a random sample of 34 players, the mean rating is ( 95.2 ). Assume the population standard deviation of the ratings is known to be ( 12.7 ). Is there enough evidence to support the claim that the mean player rating is higher than 90 ? Perform a hypothesis test, using the ( 0.05 ) level of significance. (a) State the null hypothesis ( H_{0} ) and the alternative hypothesis ( H_{1} ). [ egin{array}{l} H_{0}: \ H_{1}: end{array} ] (b) Perform a ( Z )-test and find the ( p )-value. Here is some information to help you with your ( Z )-test. - The value of the test statistic is given by ( frac{ar{x}-mu}{frac{sigma}{sqrt{n}}} ). - The ( p )-value is the area under the curve to the right of the value of the test statistic. (c) Based on your answer to part (b), choose what can be concluded, at the ( 0.05 ) level of significance, about the claim made by the studio. Since the ( p )-value is less than (or equal to) the level of significance, the null hypothesis is rejected. So, there is enough evidence to support the claim that the mean player rating is higher than 90 . Since the ( p )-value is less than (or equal to) the level of significance, the null hypothesis is not rejected. So, there is not enough evidence to support the claim that the mean player rating is higher than 90 . Since the ( p )-value is greater than the level of significance, the null hypothesis is rejected. So, there is enough evidence to support the claim that the mean player rating is higher than 90. Since the ( p )-value is greater than the level of significance, the null hypothesis is not rejected. So, there is not enough evidence to support the claim that the mean player rating is higher than 90 .

          After a lot of hard work, a studio releases a new video game. The studio believes that the game will be liked by players; in particular, the studio claims that the mean player rating, ( mu ), will be higher than 90 . In a random sample of 34 players, the mean rating is ( 95.2 ). Assume the population standard deviation of the ratings is known to be ( 12.7 ).

Is there enough evidence to support the claim that the mean player rating is higher than 90 ? Perform a hypothesis test, using the ( 0.05 ) level of significance.
(a) State the null hypothesis ( H_{0} ) and the alternative hypothesis ( H_{1} ).
[
egin{array}{l}
H_{0}: \
H_{1}:
end{array}
]
(b) Perform a ( Z )-test and find the ( p )-value.
Here is some information to help you with your ( Z )-test.
- The value of the test statistic is given by ( frac{ar{x}-mu}{frac{sigma}{sqrt{n}}} ).
- The ( p )-value is the area under the curve to the right of the value of the test statistic.
(c) Based on your answer to part (b), choose what can be concluded, at the ( 0.05 ) level of significance, about the claim made by the studio.
Since the ( p )-value is less than (or equal to) the level of significance, the null hypothesis is rejected. So, there is enough evidence to support the claim that the mean player rating is higher than 90 .

Since the ( p )-value is less than (or equal to) the level of significance, the null hypothesis is not rejected. So, there is not enough evidence to support the claim that the mean player rating is higher than 90 .

Since the ( p )-value is greater than the level of significance, the null hypothesis is rejected. So, there is enough evidence to support the claim that the mean player rating is higher than 90.

Since the ( p )-value is greater than the level of significance, the null hypothesis is not rejected. So, there is not enough evidence to support the claim that the mean player rating is higher than 90 .

$After a lot of hard work, a studio releases a new video game. The studio believes that the game will be liked by players; in particular, the studio claims that the mean player rating, ( mu ), will be higher than 90 . In a random sample of 34 players, the mean rating is ( 95.2 ). Assume the population standard deviation of the ratings is known to be ( 12.7 ). Is there enough evidence to support the claim that the mean player rating is higher than 90 ? Perform a hypothesis test, using the ( 0.05 ) level of significance. (a) State the null hypothesis ( H0 ) and the alternative hypothesis ( H1 ). [ eginarrayl H0: H1: endarray ] (b) Perform a ( Z )-test and find the ( p )-value. Here is some information to help you with your ( Z )-test. - The value of the test statistic is given by ( fracarx-mufracsigmasqrtn ). - The ( p )-value is the area under the curve to the right of the value of the test statistic. (c) Based on your answer to part (b), choose what can be concluded, at the ( 0.05 ) level of significance, about the claim made by the studio. Since the ( p )-value is less than (or equal to) the level of significance, the null hypothesis is rejected. So, there is enough evidence to support the claim that the mean player rating is higher than 90 . Since the ( p )-value is less than (or equal to) the level of significance, the null hypothesis is not rejected. So, there is not enough evidence to support the claim that the mean player rating is higher than 90 . Since the ( p )-value is greater than the level of significance, the null hypothesis is rejected. So, there is enough evidence to support the claim that the mean player rating is higher than 90. Since the ( p )-value is greater than the level of significance, the null hypothesis is not rejected. So, there is not enough evidence to support the claim that the mean player rating is higher than 90 .$

Added by Jesse L.

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