A psychological researcher has a theory that predicts women will tend to carry more cash than men. A random sample of Ersatz University students revealed that 16 females had a mean of $22.30 in their wallets with a standard deviation of $3.20, while 16 males had a mean of $17.30 with a standard deviation of $9.60.The test statistic for the researcher's hypothesis is

Answer


1.976


1.250


1.504

Answer :

Answer:

The test statistic for the researcher's hypothesis is 1.976

Step-by-step explanation:

The standard error (SE) is,

[tex]SE=\sqrt{\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2} }[/tex]

Given that,

[tex]\bar x = 22.30,S_1=3.20\\\\\bar x=17.30,S_2=9,60\\\\n_1=16,n_2=16[/tex]

[tex]SE =\sqrt{\frac{(3.20)^2}{16} +\frac{(9.60)^2}{16} } \\\\=\sqrt{(0.64+5.76)} \\\\=2.53[/tex]

The test statistic is obtained below:

[tex]t=\frac{( \bar x_1 - \bar x_2)}{\sqrt{\frac{S_1^2}{n_1} +\frac{S_2^2}{n_2} } }[/tex]

[tex]=\frac{(22.30-17.30)}{2.53} \\\\=\frac{5}{2.53} \\\\=1.976[/tex]

Therefore, the test statistic for the researcher's hypothesis is 1.976

Cricetus

The test statistics for the researcher's hypothesis will be "1.976".

According to the given question,

  • [tex]\bar x_1 = 22.30[/tex], [tex]\bar x = 17.30[/tex]
  • [tex]S_1 = 3.20[/tex], [tex]S_2 = 9.60[/tex]
  • [tex]n_1 = 16[/tex], [tex]n_2 = 16[/tex]

The standard error will be:

→ [tex]S.E = \sqrt{\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2} }[/tex]

By putting the values, we get

         [tex]= \sqrt{0.64+5.76}[/tex]

         [tex]= 2.53[/tex]

         

hence,

The test statistic will be:

→ [tex]t = \frac{(\bar x_1 - \bar x_2)}{\sqrt{\frac{S_1^2}{n_1} +\frac{S_2^2}{n_2} } }[/tex]

     [tex]= \frac{22.30-17.30}{2.53}[/tex]

     [tex]= \frac{2}{2.53}[/tex]

     [tex]= 1.976[/tex]

Thus the above answer is correct.

Learn more about psychological researcher here:

https://brainly.com/question/13100685

Other Questions