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Consider a random sample of 10 students having 116 cm as mean height and standard deviation as 9.798 cm. If the suggested mean height of the students population is 110 cm then the t-test statistic of the sample is:

[given 109.798=0.3227\frac{\sqrt{10}}{9.798} = 0.3227]

Solution

✅ Correct Option: 1

Given:

Sample size n=10n = 10 students

Sample mean xˉ=116\bar{x} = 116 cm

Standard deviation s=9.798s = 9.798 cm

Population mean μ0=110\mu_0 = 110 cm

109.798=0.3227\frac{\sqrt{10}}{9.798} = 0.3227


The t-test statistic formula is:

t=xˉ−μ0s/nt = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}


Substituting the values:

t=116−1109.79810t = \frac{116 - 110}{\frac{9.798}{\sqrt{10}}}

t=69.79810t = \frac{6}{\frac{9.798}{\sqrt{10}}}


Rewriting the denominator:

t=6×109.798t = 6 \times \frac{\sqrt{10}}{9.798}

Using the given value:

t=6×0.3227t = 6 \times 0.3227

t=1.9362t = 1.9362

t≈1.93t \approx 1.93


Therefore, the t-test statistic of the sample is 1.931.93.

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