Null hypothesis "H_0:\\;\\mu=78."
Alternative hypothesis "H_a:\\;\\mu\\neq78."
Test statistic: "t=\\frac{84-78}{\\frac{16}{\\sqrt{17}}}=1.55."
P-value: "p=0.1407."
Since the P-value is greater than 0.05, fail to reject the null hypothesis. The difference after 6 weeks of instruction is consistent with what is likely under conditions of chance above.
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