# baseball fun

#### saige30

##### New Member
The Atlanta Braves have a batting avg of mew = .287 with a s = .025. One of the sports trainers asked the batters to start wearing amber colored sun glasses to cut down on the glare and help them to see better. In the last 16 games, their batting avg has increased to xbar = .299. Using an alpha of .05 did the team batting avg increase significantly when they started wearing amber colored glasses?

I'm confused...but I think it should set up like this...
Ho: mew > .287 vs Ha: mew < .287

I am having serious issues, I got my t to = 1.92...

not knowing anything about baseball doesn't help to know if I am even close to correct...

any help is greatly appreciated!

#### Dragan

##### Super Moderator
The Atlanta Braves have a batting avg of mew = .287 with a s = .025. One of the sports trainers asked the batters to start wearing amber colored sun glasses to cut down on the glare and help them to see better. In the last 16 games, their batting avg has increased to xbar = .299. Using an alpha of .05 did the team batting avg increase significantly when they started wearing amber colored glasses?

I'm confused...but I think it should set up like this...
Ho: mew > .287 vs Ha: mew < .287

I am having serious issues, I got my t to = 1.92...

not knowing anything about baseball doesn't help to know if I am even close to correct...

any help is greatly appreciated!

Your t statistic is correct i.e. t = 1.92.

Simply compare it to the critical value of t for a one-sided test:

tcrit = 1.753 where your degrees of freedom are 15 for alpha = 0.05.

Because 1.92 is greater than 1.753 than reject the null hypothesis that you provided.

Note: On the other hand, keep in mind that if you provided a null hypothesis of Mu = .287 vs the alternative of Mu does not equal .287 then you would have to consider both sides of the distribution.

In this case you would fail to reject the null because tcrit = 2.131 which less than 1.92.